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RECOMMENDATIONS. 

Whilst  this  work  was  passing  through  the  press,  the  sheets  were 
submitted  for  inspection  to  several  teachers  and  gentlemen  capable 
of  judging  of  the  merits  of  the  work,  and  the  following  testimonials 
have  been  received.  The  reader  will  bear  in  mind  that  these  no- 
tices are  all  from  'practical  Instructors,  who  are  now,  or  have  been, 
until  recently,  engaged  ii  teaching;  and  also,  that  the  majority  of 
them  have  been  furnished  wkhout  solicitation,  and  may  therefore  be 
regarded  as  in  every  sense  of  the  word  impartial. 

From  the  Rev.  J.  H.  Good,  A.  M.,  of  Columbus,  formerly  Rector 
of  the  Preparatory  Department  of  Marshall  College,  Pa.,  and  for 
many  years  a  practical  teacher, 

Mr.  C.  H.  Mattoon — Dear  Sir  : 

I  have  examined  your  "  Common  Arithmetic  " 
with  considerable  care,  having  perused  nearly  the  whole  of  it  as  it  passed 
through  the  press.  It  is  an  excellent  practical  work.  From  its  arrangement, 
its  fullness  of  Rules  and  Examples,  its  clearness  of  explanation — above  all,  its 
eminently  practical  character,  I  regard  it  as  peculiarly  well  adapted  to  the  wants 
of  that  large  class  of  persons,  who  usnally  pass  through  but  one  work  on 
Arithmetic  preparatory  to  their  entrance  upon  public  life.  Every  teacher  can 
at  once  perceive,  by  examination,  how  well  it  is  suited  to  the  wants  of  our  Com- 
mon Schools.  I  can  cordially  recommend  it  to  the  careful  examination  of 
teachers.  Yours, 

Respectfully, 

J.  H.  GOOD. 


From  the  Rev.  E,  W.    Clark,  formerly  an  Agent  for    Granville 
College,    Ohio, 

Genoa,  Delaware  Co.,  0.,  December  24th,  1849. 
Mr.  C.  H.  Mattoon — Sir  : 

I  have  examined  those  sheets  of  your  Arithmetic 
which  you  sent  me,  and  would  say  that  I  am  well  pleased  with  the  same.  I 
consider  the  definitions  accurate — the  arrangement  natural  and  scientific — the 
introduction  and  application  of  the  later  improvements  in  this  science  judicious 
— the  rejection  of  much  old  and  useless  matter,  and  the  introduction  in  its 
stead  of  a  large  number  of  practical  Rul«s,  an  improvement — and,  in  short, 
considering  its  clearness  and  simplicity  of  style,  together  with  its  thoroughness 
and  comprehensiveness  of  subjects,  and  its  full  and  extensive  application  to 
practical  questions,  I  would  say  that  I  think  it  better  adapted  to  meet  the  wants 
bo'h  of  our  Common  and  High  Schools,  than  any  other  work  of  the  kind 
with  which  I  am  acquainted. 

"  I  am  respectfully  yours, 

'•  E.  W.  CLARK. 


RECOMMeN  D  AT  [ONS . 

Extract  from  a  letter  from  Mr.  Barton  Moore,  a  thorough  practical 
instructor,  near  Sunhery,    0- 

"  I  received  the  copies  of  your  Arithmetic  which  you  sent  me,  and  in 
those  I  have  examined,  I  believe  the  principles  of  numbers  can  be  more  easily 
understood  by  the  youth,  and  the  method  of  explanation  is  far  superior  to  any 
with  which  I  am  acqudinted. 

*'  Yours,  &c., 

«B.  MOORE." 

From  Mr.  S.  L.  Wallace,  a  practical  teacher  near  Kingston,  Ross 

county,  Ohio. 

Kingston,  Ross  Co.,  O.,  Dec.  28,  1849. 

Mr.  C.  H.  Mattoon  : — I  received  those  sheets  of  your  book,  which  you 
were  kind  enough  to  send  me,  and  after  a  careful  perusal  of  them,  I  am  satis- 
fied that  the  work  is  better  adapted  to  the  use  of  common  schools,  and  acade- 
mies, than  any  other  work  of  the  kind  with  which  I  am  acquainted. 

1st.  Because  it  comprises  more  useful  and  practical  matter  than  any  other 
work  on  Arithmetic  extant. 

2nd.  On  account  of  its  perfect  adaptation  to  the  purposes  of  instruction, 
together  with  its  thoroughly  progressive  character ;  one  subject,  and  one  difficulty 
only,  at  a  time,  being  presented,  and  that  made  perfectly  familiar  before  pro- 
ceeding to  the  next  step. 

3d.  The  style  is  c'ear  and  easy,  thus  fitting  it  for  the  lawer  as  well  as  for 
the  higher  classes  of  learners. 

4th.  Although  it  is  plain  and  simple,  it  is  also  full  and  comprehensive,  and 
the  continual  reference  to  preceding  principles,  &c.,  together  with  many  orig- 
inal and  important  remarks  interspersed  throughout  the  work,  form  a  material 
help  to  the  learner. 

5th.  In  the  precision  and  accuracy  of  the  definitions  ;  in  the  natural  and 
scientific  arrangement  of  subjects;  in  the  careful  and  judicious  introduction 
and  application  of  the  later  improvements;  in  the  clear  and  lucid  explanation 
of  hard  and  difficult  subjects;  and  in  th«  tersenesss  and  comprehensiveness  of 
the  rules,  it  far  excels  any  other  work  upon  this  subject  that  has  yet  fallen 
under  my  observation,  and  supplies  an  important  desideratum  long  wanted  in 
our  Arithmetics. 

6th.  The  rejection  of  much  old  and  useless  matter,  and  the  introductioin  of 
many  rules  applicable  to  practical  business,  which  heretofore  have  receivd  but 
little  or  no  attention,  speak  well  in  its  favor. 

In  short,  I  consider  it  a  work  of  sterling  merit,  and  superior  excellence,  and 
decidedly  preferable  to  a  majority  of  the  trash  that  at  this  day  is  being^  crowded 
upon  our  schools.  I  remain,  yours  respectfully, 

S.  L.  WALLACE. 

JFrom  Mr.  Wm.  C.  Gildersleeve,  an  old  and  experienced  teacher 
near  Chillicothe,  Ohio. 

Chillicothe,  0.,  Dec.  29th,  1849. 
Mr.  Charles  H.  Mattoon — Sir: — I  have  carefully  examined  your  work 
entitled  "Common  Arithmetic"  and  unhesitatingly  say,  that  in  my  opinion,  it 
is  far  superior  to  any  other  work  of  the  kind,  I  have  yet  perused.  The  rules, 
(some  of  which  are  new  to  me)  are  better  calculated,  and  more  nearly  related  to 
the  nature  and  uses  of  mathematical  computations,  in  ail  ordinary  business,  than 
those  usually  given.  Therefore,  on  account  of  its  simplicity,  comprehensiveness 
and  intrinsic  merit,  I  would  reccommend  it  to  the  Schools  and  Academies  of 
our  Western  country;  Yours  respectfully, 

WM.  C.  GILDERSLEEVE. 


ANALYTIC  SERIES,  NO.  2. 


COMMON  ARITHMETIC. 


UPON  THE   ANALYTIC  METHOD  OF  INSTRUCTION 


ALSO  :    THE 


PRINCIPLES    OF    CANCELATION 


AND  OTHER  MODERN  IMPROVEMENTS. 


ILLUSTRATED    AND    APPJ.IED. 


The  wliole  made  simple  and  easy  by  numerous  practical  examples. 


DESIGNED  FOR  THE  USE  OF  SCHOOLS. 


BY  CHARLES  H.  MATTOOX. 


There's  nothing  so  hard,  but  search  will  find  it  out.' 


t: 


EAM    PKESS    OF   S.    MEDARY, 


1860.1 

ivi^?'0538  ' 


Entered,  according  to  Act  of  Congress,  in  the  year  1849, 

By  CHARLES  H.  MATTOON, 

In  the  Clerk's  Office  of  the  District  Court  of  Ohio. 


PREFACE. 

In  presenting  this  treatise  to  the  public,  the  Author  does  not 
think  it  necessary  either  to  give  his  reasons  for  so  doing,  or  to  apolo- 
gise for  its  appearance.  For  the  former,  the  student  cares  but  little, 
and  for  the  latter,  still  less. 

Neither  will  he  enumerate  what  he  considers  to  be  its  exc  ^llencies 
being  convinced  that  if  it  really  possesses  merit,  its  worth  will  be 
discovered  by  others,  without  his  blazing  it  abroad. 

He  will,  however,  present  a  few  of  its  most  prominent  character- 
istics for  the  consideration  of  the  public,  and  leave  it  for  them  to 
decide  whether  they  add  to,  or  detract  from  the  merits  of  the 
work. 

1,  Each  section  commences  with  a  number  of  MerUcU  JSJxercises, 
in  order  to  give  the  pupil  an  idea  of  the  subject  before  he  is  burdened 
with  learning  abstract  rules  and  principles. 

2,  Dsjinitions  are  then  given,  and  the  principles  demonstrated 
and  explained  separately ;  the  first  being  proved  from  self-evident 
propositions,  and  those  following,  by  those  already  explained.  In 
no  case  is  one  principle  used  to  explain  another,  until  it  has  itself 
been  explained, 

3.  The  principles  are  then  summed  up  into  a  general  rule,  for  ref-  * 
erence  and  review.     After  the  rule  follow  a.  number  of  Exercises 

for  the  Slate,  and  these,  as  well  as  the  Mental  Exercises  are  chiefly 
of  a  practical  nature,  and  strictly  confined  to  the  principles  already 
explained. 

4.  Those  Rules  most  used  in  practical  business  have  received 
more  particular  attention,  than  those  of  less  importance. 

6.  Pounds,  shillings  and  pence  are  rejected  entirely,  and  their 
places  supplied  with  Federal  Money.  If  we  have  a  national  curren- 
cy, why  not  adopt  it  exclusive  of  all  others? 


It  preface. 

6.  The  tables  of  Weights  and  Measures  have  been  prepared  with 
direct  reference  to  existing  laws  and  present  uses. 

7.  The  subject  of  Cancelation  is  explained,  and  the  method  of  ap- 
plying its  principles  to  business  distinctly  shown.  Unlike  some 
modern  writers,  however,  the  Author  does  not  make  it  the  chief  of 
all  rules.  Other  modern  improvements  are  also  illustrated  and  ap- 
plied. 

8.  The  Rules  of  the  Cu  he  Root,  Progression,  Alligation,  <&c,,  are 
,  omitted  for  three  reasons  :  1st.  They  seldom  if  ever  occur  in  com- 
mon business,  and  those  who  have  occasion  to  use  them  in  their 
particular  pursuits,  are  generally  men  of  science,  who  are  acquain- 
ted with  the  higher  departments  of  mathematics,  in  which  they  are 
explained  more  clearly  and  satisfactorily  than  they  can  be  in  an  ele- 
mentary work  on  Arithmetic.  2nd.  Their  place  can  be  supplied 
with  rules  of  more  importance  ;  that  is,  rules  that  frequently  oc- 
cur in  ordinary  transactions.     3d.  They  increase  the  size  of  a  book, 

and  the  labors  of  the  learner  unnecessarily.  The  Author  has  yet 
to  be  convinced  of  the  necessity  of  burdening  the  mind  with  ab- 
struse and  metaphysical  questions,  or  tedious  and  complicated  dem- 
onstrations, which  in  nine  cases  out  of  ten  can  never  be  of  any 
practical  utility  to  the  pupil. 

9.  The  rules  in  Mensuration  are  demonstrated,  with  but  few  ex- 
ceptions. The  language,  however,  is  chosen  more  for  perspicuity 
than  for  mathematical  precision. 

10.  The  language  adopted  throughout  the  work,  is  simple,  terse 
and  definite.  Childishness  and  puerility  are  avoided  on  the  one 
hand,  as  much  as  complication  and  mystery  on  the  other.  The  aim 
has  been  to  unite  brevity  with  perspicuity. 

11.  The  arrangement  of  subjects  is  such  as  seems  the  most  con- 
sistent with  the  natural  order  of  the  sciences.  Therefore  Fractions 
follow  immediately  after  the  Fundamnetal  Rules;  because,  1st,  they 
partake  of  the  nature  of  Division;  and  2nd.  They  are  frequently 
used  in  Compound  Numbers;  hence,  it  is  necessary  to  understand 
Fractions  before  the  operations  in  Compound  Numbers  can  be 
thoroughly  understood.  Also,  Federal  Money  is  placed  with  Deci- 
mal Fractions,  because  the  same  rules  are  applicable  to  both,    since 


PREFACE.  V 

they  are  based  on  the  same  scale  of  notation.  Likewise,  Commis- 
sion, Insurance,  Interest,  dbc.  are  placed  under  PercentagCj  upon  the 
principles  of  which  they  are  based. 

12.  The  analogies  andi  relations  oi  numbers  are  clearly  pointed 
out  and  illustrated  ;  and  from  these,  the  scholar  is  taught  to  reason 
— to  think  for  himself.  He  is  required  to  believe  no  theory  without 
proof  and  to  take  no  principle  without  demonstration.  He  is  every 
where  taught  that  there  is  no  guess  work  about  results — that  all  ques- 
tions in  Arithmetic  can  be  solved  upon  true  principles — that  there  is 
no  uncertainty  about  mathematical  conclusions — in  short  that  every 
theory  or  principle  connected  either  directly  or  indirectly  with  math- 
ematics, that  is  not  capable  of  demonstration,  must  be  erroneous. 
Thus  he  is  led  to  investigate,  and  examine — to  exercise  his  reasoning 
faculties — to  work  iniellectually,  instead  of  mechanically — and  to  be- 
lieve a  theory  or  principle  because  it  is  clearly  and  conclusively  pro- 
ven, and  not  merely  because  "the  book  says  so.'' 

Such  is  a  brief  outline  of  the  present  work.  We  have  endeavored 
to  present  every  principle  that  a  business  man  may  have  occasion 
to  use,  in  a  clear  and  systematic  manner,  and  also,  to  prepare  it  in 
such  a  way  as  to  be  adapted  to  precede  the  study  of  Algebra,  and 
the  higher  branches  of  mathematics.  Our  object  has  not  been  to 
produce  an  abstruse  and  metaphysical  treatise,  but  simply  to  present 
a,  plain,  practical,  and  comprehensive  work,  which  should  meet  the 
wants  of  our  business  men,  and  arithmetical  students.  We  have 
rejected  many  old,  useless,  and  obsolete  rules,  and  supplied  their 
place  with  such  modern  improvements  as  were  deemed  of  any  im- 
portance. 

In  short,  we  have  endeavored  to  say  everything  that  was  neces- 
sary, and  no  more.  The  work  is  designed  for  use,  and  not  merely 
for  show.  It  was  prepared  by  the  Author  whilst  engaged  as  a 
teacher,  and  is  the  result  of  arduous  labor,  and  hard  study,  aided 
by  practical  observation  and  experience  in  the  school  room. 

It  may  be  proper  to  remark  here,  that  this  is  designed  as  the  sec- 
ond book  of  a  series,  now  in  course  of  preparation  by   the  Author, 
The  first  work,  entitled  '^ Arithmetic;  Mental  and  Practical"  for  be- 
U 


VI  PREFAeE- 

ginners,  is  in  progress  and  will  be  published  probably  some  time  the 
ensuing  season. 

How  successfully  our  work  has  been  executed,  time  alone  will  de- 
cide. For  the  present  we  submit  it  to  teachers  and  scholars  for  ex- 
amination, convinced  that  it  must  stand  or  fall,  upon  its  own  merits 
alone.  And  should  it  be  found  to  lessen  the  labors  of  teachers,  or 
to  promote  the  intellectual  attainments  of  scholars,  its  highest  aims 
will  be  accomplished. 

THE  AUTHOR. 


CONTENTS 


Page 

Suggestions  for  Teachers xvi 

SECTION  T. 

Arithmetic  defined 1 

Notation 1 

"     the  Boman  method 2 

"     the  ^raSic  method 3 

Numeration 4 

Numeration  Table 5 

Exercises  in  Numeration 7 

Exercises  in  Notation:  _  , ,       8 

SECTION  11. 

Addition — Mental  Exercises 9 

Definitions — Addition  Table 10 

Proof  of  Addition 11 

Carryingin  Addition 12 

General  Bale — Exercises  for  the  slate 13 

SECTION  III. 

Subtraction — Mental  Exercises — Definitions 16 

Subtraction  Table 17 

Proof  of  Subtraction 18 

Borrowing  and  carrying  in  Subtraction  illustrated 19 

General  Rule — Exercises  for  the  slate 20 

SECTION  IV. 

Multiplication — Mental  Exercises — Definitions, 23 

Multiplication  Table 24 

Proof  of   Vlultiplicition 26 

Methai  of  carrying  ill.jstrated^ — Rule  when  the  multiplier  is 

less  than  12 27 

General  Rule — Exercises  for  the  slate 30 


Viii  CONTENTS. 

Page- 
Contractions  in  Multiplication — When  the  multiplier  is  10,  100, 

1000,  &c. — To  multiply  by  a  Composite  number 32 

When  there  are  ciphers  at  the  right  of  either  or  both  factors  _  34 

m 

SECTION  V. 

Division — Mental  Exercises 36 

Definitions — Division  Table 37 

Proof  of  Division — Proof  of  Multiplication  by  Division 39 

Remarks  respecting  the  Remainder 40 

Method  of  carrying  ilUustrated 41 

Rule  for  Short  Division 42 

Exercises  for  the  Slate 43 

Lq;tig  Division 44 

Rule  for  Long  Division — Exercise  for  the  Slate 46 

Contractions  in  Division — To  divide  by  10,   100,  1000,  &c-_  48 

To  divide  by  a  composite  number 49 

When  there  are  ciphers  at  the  right  of  the  divisor 60 

SECTION  YL  J 

I^ECAPiTULATiON — Remarks  and  inferences — Notation 51 

numeration — Addition 62 

Subtraction 63 

Multiplication  __^ 64 

Proof  by  casting  out  the  9's 65 

Division 67 

Proof  by  casting  out  the  9's 69 

Principles  deduced  from  Division 60-62 

Contractions  and  Abbreviations — To  multiply  by  26.      By  £0. 

By  126 62 

—When  part  of  the  multiplier  is  a  factor  of  another  part 62 

-^By  aid  of  exponents 63 

To  divide  by  26 64 

— By  126.     When  there  is  a  remainder  after  dividing  by  sev- 
eral divisors,  to  find  the  true  remainder 66 

To  contract  operations  in  Long  Division && 

SECTION  YII. 

Cancelation — Definitions  and  Illustrations 67 

General  Rule 70 

Exercises  for  the  slate 71 

(jf.rfiatest  Common  Divisor 72 

Rvfle — for  two  numbers , 73 

— For  more  th^n  two  numbers — two  rules 74 

Le,ast  Common  'Multiple 76 

]^ule — Exercises  for  the  slate_  .* . ,i--. ..-..--.  80 


CONTENTS.  IX 

Page. 

SECTION  VIII. 

Fractions — Mental  Exercises '.  8 1 

Common  Fractions — Definitions 85 

Fundamental  Propositions 87 

Reduction   of  Fractions — To  reduce  fractions    to  its   lowest 

terms 90 

— An  Trnproper  fraction  to  a  whole  or  mixed  number 91 

— A  whole  or  mixed  number  to  an  improper  fraction 92 

— A  compound  fraction  to  a  simple  one  __• 93 

— Ditto,  by  cancelation 94 

To  reduce  fractions  to  their  least  common  denominator 94 

General  Mules  for  the  Reductipn  of  Fractions 96 

Exercises  for  the  slate 97 

Addition  and  Subtraction  of  Fractions — Mental   Exercises 98 

General  Mule — Exercises  for  the  slate 100 

Multiplication  of  Fractions — Mental  Exercises 102 

Exercises  for  the  slate — To  multiply  a   fraction  and   a  whole 

number  together 103 

To  multiply  a  whole  number  and  a  mixed  number  together.  _  105 

— Fractions  together 106 

— Mixed  numbers  together 107 

General  Mules  for  the  Multiplication  of  Fractions 108 

Exercises  for  the  slate ,__^ 108 

Division  of  Fractions 109 

Mental  Exercises — Exercises  for  the  slate — To  divide  a  frac- 
tion by  a  whole  number 110 

To  divide  a  whole  number  by  a  fraction 111 

— One  fraction  by  another _. 113 

Method  of  proceeding  when  mixed  numbers  occur 114 

Complex  fractions 116 

To  change  a  fraction  to  any  required  denominator 118 

General  Mules  for  the  Division  of  Fractions 120 

Exercises  for  th^  slate 121 

M  iscellaneous  exercises  for  the  slate 123 

Decimal  Fractions — Definition  ?,  iS:c 124 

Decimal  Numeration  Table 125 

Numeration,  and  Notation  of  Decimals 127 

Federal  Money — Table 128 

To  read  sums  in  Federal  Money 130 

To  write  sums  in  Federal  Money 131 

Meduction  of  Federal  Money,  dc 131 

Exercises  for  the  slate— Reduction  of  Decimal  Fractions 132 

To  cl)  ange  a  Decimal  to  a  Common  Fraction 132 

To  change  a  Common  fraction  to  a  Decimal 133 


X  CONTENTS. 

Addition  and  Subtraction  of  Decimals  and  Federal  Money 1 35 

Rule -  137 

Exercises  for  the  slate 138 

Maltiplication  of  Decimals  and  Federal  Money 139 

Rule — Exercises  for  the  slate 140 

Method  of  buying  and  selling  articles  by  the  100  or  1000 141 

Division  of  Decimals  and  Federal  Money 142 

Rule -- 143 

Exercises  for  the  slate 144 

Miscellaneous  exercises  for  the  slate 145 

Bills,  Accounts,  &c 146 

SECTION  IX. 

Compound  IS'tjmbers — Definitions 148 

Reduction 149 

Weights — Troy    Weight 150 

Exercises  in  Troy  weight 151 

Rales  for  Reduction,  Ascending  and  Descending 152 

Avoirdupois  Weight __    152 

Exercises  in  Avoirdupois  Weight 153 

Apothecaries  Weight 153 

Exercises  in  Apothecaries  Weight 154 

Measures  OF  Capacity — Dry  Measure . 154 

Exercises  in  Dry  Measure 155 

Wine  Measure 156 

Exercises  in  Wine  Measure 156 

Ale  or  Beer  Measure , 157 

Exercises  in  Ale  or  Beer  Measure 157 

Measures  -of  Extension — Long  Measure 158 

Exercises  in  Long  Measure 159 

Cloth  Measure 159 

Exercises  in  Cloth  Measure 160 

Land  or  Square  Measure 161 

Exercises  in  land  or  Square  Measure _._  162 

Solid  or  Cubic  Measure . 164 

Exercises  in  Solid  or  Cubic  Measure  . -. 165 

Time 166 

Exercises  in  Time 167 

Circular  Measure  or  Motion 168 

Exercises    in  Circular  Measure  or  Motion 169 

Miscellaneous  Exercises  in  Reduction 169 

Fi  actions  of  Compound  Numbers 174 

To  reduce  higher  to  lower  fractions — Rule 174 

To  reduce  fractions  to  integers,  and  integers  to   fractions — 

Rule _ 175 


CONTENTS.  XI 

Page. 

To  reduce  decimals  to  fractions  and   fractions  to  decimals 177 

Addition  of  Compound  lumbers 179 

Rule 180 

Exercises  in  Addition  of  Compound  Numbers 181 

Subtraction  of  (JcmipouTid  Numbers 183 

Rule 184 

Exercises  in  Subtraction  of  Compound  Numbers 185 

MuW plication  and  Division' of  Compound  Numhers 187 

Rules . 187 

Exercises  in  Multiplication  and  Division  of  Compound  Num- 
bers  . 188 

Diflference  in  time  of  two  places — Rule ^^ 193 

SECT[ON  X. 

Exchange — Definitions,  <&c . 195 

Table  of  Foreign  coins 196 

Rule — To  change  foreign  coins  to  Federal  Money 196 

Rule — To  change  Federal  Money  to  Foreign  Coins 197 

SECTION  XL 

Analysis. — Definitions,  Sc. 198 

Mental  Exercises  in  Analysis 199 

Exercises  for  the  Slate 201 

Practice — Definitions,  &c 207 

Exercises  in  do 208 

Barter,  Definition  ajid  Exercises. 212 

Alligation — Medial 213 

Porportional  Division 214 

Exercises  in  Analysis 215 

SECTION  XII. 

Ratio  and  Proportion — Definition  of  iSa^io 219 

Mental  Exercises  in  do 220 

Rules  and  Exercises  in  Ratio .  220 

Simple  and  Compound  Ratio 223 

Proportion — Definition,  &c 224 

Difference  of  Ratio  and   Proportion . 225 

Extremes  and  Means  in  Proportion 225 

Rules  in  Proportion 226' 

Mental  Exercises 227 

Simple  Proport  on,  Definitions,  &c 227 

Rules  for  working  questions  in  Simple  Proportion 229 

General  Rule  for  Simple  Proportion 232 

Exercises  for  the  slate 233 


Xll  CONTl'NT.««. 

Page. 

Compound  Proportion,  Definitions,  &c 235 

Solution  of  questions  in  do 236 

How  to  prove  Compound  Proportion 237 

General  Eidejor  Compound  Proportion 238 

Exercises  for  the  slate  n  do . 239 

Partnership,  Definition,  Solutions,  &c 240 

Rule  for  Partner sh  ip,  Exercises 241 

Bankruptcy,  Exercises 242 

General  Average,  Exercise 242 

SECTION  XIII. 

Percentage — Definitions „ 244 

Mental  Exercises 245 

Rate  for  Percentage,  Exercises 247 

Commission  and  Insurance 249 

Exercises  in  C'ommmiow 250 

Exercises  in  Insurance 251 

Profit  and  Loss,    Definitions 254 

Exercises  in  do . 255 

Stocks,   Definitions " 260 

Exercises  in  do. 261 

Brokage,  Definitions,  Exercises 262 

Duties,  Definitions 262 

Specific  Duties,  Definitions 263 

Exercises  in  do. 264 

Ad  Valoram  Duties,  Definitions _. '-  264 

Taxes,  Definitions 265 

General  Rule  for  Taxes 266 

Table  for  Taxes 267 

Interest,  Definitions  and  Illustrations 2j8 

Simple  Interest,  Illustration 269 

Rule — Exercises 270 

Rule  for  years  and  months 27 1 

Rule  for  days 272 

General  Rule  for  Interest 272 

Table  for  Interest _" 273 

Exercises  in  Interest 274 

Interest  at  any  rate — Rule 276 

"Second  method  for       **     277 

Rule  for  Interest  on  Notes  and  Bonds 279 

Vermont  Rule         "          "          "            281 

Connecticut  Rule     "          **          **            282 

Exercises  in  the  above 283 

Problems  in  Interest _.- _^ _  2o5 


UONTLNT'^.  XIU 

To  find  the /71/eres/,  To  find  the  raie 285 

To  find  the  time  _._ 286 

To  find  the  Principal 287 

To  find  the  Principal  and  Interest 288 

Discount,  Definitions 289 

Exercises  in • ^^^ 

Banking,  Definitions 29 1 

Exercises  in 2.2 

Equation  of  Payments^  illustrations 293 

Rule  for  finding  equoXed  Time 294 

Compound  Interest,  Illustrations 296 

Table  for  Compound  Interest 297 

Exercises  in             ** 298 

SECTION  XIV. 

Duodecimals,  Table,  Multiplication  of 299 

Rale  for  Multiplication  of  Decimals, 30 1 

Painters'  Powers  and  Plasterers'    Work •  30 1 

Masons'    Work , 303 

SECTION  XV. 

Powers  and  Roots,  Definitions 303 

Involution,   Definitions,  Exercises 304 

Rule  for  Involution,  Exercises 305 

Involution  of  Fractions , 306 

Evolution,  Definitions 308 

Mental  Exercises,  Illustrations 309 

Extraction  of  the  Square  Root,  Illustrations 311 

Illustrations,  Diagrams, 313 

Rule  for  Extracting  the  Square  Root 315 

Exercises  in                     "           "         . 317 

Rule  to  Extract  the  Square  Root  of  Fractions 318 

Triangles,  Illustrations 319 

Mean  Proportionals 321 

Circles,  Definitions,  Illustrations 323 

SECTION  XVI. 

Mensuration,  Mensuration  of  Surfaces, 324 

''Area  of  Triangles,  Rule 325 

Area  of  Trapezoids,  Rule 326 

^  Area -of  Circles  and  Polygons 327 

'Circumscribed  and  Inscribed  Polygons 328 

U^Quadrature  of  the  Circle _  329 

If  Satio  of  the  Diameter  to  Circumference -i£ls^S:*i,i 330 


XIV  ©ONTENTS. 

To  find  the  Circumfereyice,  Diameter 330 

Rule  for  finding  the  Area  of  the  Circle ^ 331 

Area  of  a  Square  inscribed  in  a  Circle ««.._  332 

Dimensions  of  the  Circumscribed  Circle 336 

Mensuration  of  Solids,  Prisms,  Area  of  a  Piism 338 

To  find  the  solidity  of  a  Prism 339 

Cylinders,  Definitions ^ 339 

To  find  the  solidity  of  a  Cylinder .^  340 

Boilers,  To  find  the  contents  of ,__  340 

To  find  the  contents  of  a  cistern 34] 

Parallelopiped,  to  find  the  surface  and  solidity 342 

-PyramJ6^5,  Definitions 342 

To  find  the  surface  of  a  Pyramid  or  Cone 343 

To  find  the  surface  of  the  Frustrum  of  a  Pyramid  orCone  _>  344 

To  find  the  solidity  of  a  Pyramid  or  Cone 344 

To  find  the  solidity  of  the  Frustrum  of  a  Pyramid  or  Cone  _  _  345 
Spheres  or  Globes.     To  find  the  surface  of   a  Sphere  or 

Globe 346 

To  find  the  solidity  of  a  Globe  or  Sphere « 346 

Tonnage  of  Vessels,  Definitions 346 

Rule  for  calculating  tonnage ^ 347 

To  find  the  solidity  of  Irregular  Bodies ._.__  348 

Quaging.     To  find  the  number  of  bushels  a  vessel  will  con- 
tain    348 

To  find  the  number  of  wine  and  beer  gallons  in  a  cask 349 

SECTION  XVII. 

Philosophical  and  Miscellaneous   Questions  . .  . .. 350 

The  Mechanical  Powers,   Definitions _ „  350 

The  Lever ^  Definitions  and  Rules _  351 

The  Wheel  and  Axle,  Definitions  and  Rules 352 

The  Pully,  Definitions  and  Rules 352 . 

The  Inclined  Plane,  Definitions  and  Rules ,  353 

The   Wedge,  Definitions  and  Rules 354 

The  Screw,  Definitions  aud  Rules 354 

Machinery,  Defininitions  and  Rules .  355 

Book  Keeping,  D  efinitions  and  Forms 356 

Book  Keeping  for  drovers  and  farmers 357 

Forms.     Form  of   Orders — Oi  Meceipts 358 

Form  of  Due-bills— Of  Mies 359 

Remarks  on  Notes ..,,     _.  356 

Miscellaneous  Problems  and  Pules ^ 360 

To  find  the  heio-hth  of  an  object — its  distance 261 

To  find  the  time  a  time  a  body  has  been  falling — Its  velocity.  361 

To  find  the  space  fallen  through   362 


PREFACE. 

Problem  relating  to  the  right-angled  triangle 362 

To  measure  standing  timber . 362 

To  find  the  temperature  of  any  day — De  Witt's  Rule 363 

Various  Problems 363 

Analytic  Questions I 365 

SECTION  XVIII. 

MiSCELLAiiEOUS    EXAMPLES    FOR    THE     SlATK 368 

Errata 387 


■7 

INTPvODUCTORY    REMARKS 

FOR  TEACHERS. 


There  are  at  least  three  different  departments  of  science  with 
which  every  person  should  be  familiar,  and  they  are,  Reading, 
Writing,  and  Arithmetk.  These,  in  consequence  of  their  universal 
application  to  all  kinds  of  business,  might  with  propriety  be  termed 
the  Golden  Branches  of  Science.  Other  departments  may  be  desi- 
rable for  ornament,  or  necessaiy  in  some  particular  profession  or 
occupatioii,  or  useful  in  storing  the  mind  with  general  information  : 
but  there  are  none  that  will  apply  so  universally  to  all  classes  of 
individuals,  and  to  every  department  of  business,  as  the  three 
branches  above  named. 

Such  being  the  case,  how  necessary  is  it  that  they  should  be 
taught  correctly,  thoroughly,  and  completely :  that  the  learner 
should  be  taught  correct  theories,  correct  prmcijoles^  and  the  correct 
'method  of  applying  these  theories  and  principles  to  practical  use — '■ 
in  short,  that  he  should  be  taught  to  work  intellectually,  instead  of 
iiechanically . 

in  Arithmetic,  especially,  is  it  necessary  to  be  particular,  as  much 
is  often  at  stake  by  a  single  operation  ;  and  the  greater  the  risk,  the 
more  the  need  of  care. 

In  teaching  Arithmetic,  many  things  are  to  be  considered,  the 
principal  of  which  are — 

Isfc.     The  age  of  the  pupil ;  and 

2nd.     His  natural  abilities  for  receiving  instructions. 

With  young  pupils,  the  subject  under  consideration  should  be 
presented  in  clear,  simple,  and  definite  language — free,  on  the  one 
hand,  from  childishness  and  puerility,  and  on  the  other  hand  from 
complication   and  mystery.     If  possible,  technical  terms  should  be 


rNTRODUCTORY    REMARK*".  XVll 

avoided  ;  but  if  it  is  absolutely  necessary  to  make  use  of  any,  they 
should  be  f^dly  and  thoroughly  explained.  As  he  advances,  addi- 
tional principles  may  be  presented  to  his  view,  and  new  theories  ex- 
plained for  his  consideration  ;  and  thus  he  may  be  led  on,  step 
by  step,  gradually  and  thoroughly,  until  he  has  mastered  the  first 
;  elementary  rudiments  of  mathematical  science,  and  is  able  to  grap  - 
^ple  successfully  with  the  more  abstruse  and  difficult  propositions. 

As  his  mind  enlarges,  and  his  intellect  expands,  and  his  ability 
for  receiving  instruction  increases,  the  explanatory  language  may 
be  proportionably  altered,  and  some  of  the  technicalities  of  the  sci- 
ence may  be  introduced  ;  the  pupil  has  now  learned  the  first  rudi- 
ments and  is  able  to  think  and  reason  to  some  extent  for  himself, 
and  needs  not  the  particular  demonstration  of  every  simple  problem, 
or  illustration  of  every  common  term. 

As  he  advances  still  farther,  the  illustrations  may  be  in  strict  sci  - 
eniijic  language,  and  he  may.  be  required  to  investigate  theories,  ex- 
amine principles,  demonstrate  abstruse  propositions,  unravel  difficult 
questions,  or  solve  knotty  problems— to  prove  why  a  theory  or 
statement  is,  or  is  not  correct,  and  in  short,  to  reason,  think,  exam- 
ine, and  demonstrate  a  truth  by  his  own  investigating,  powers,  inde- 
pendent of  the  **  say  so  "  of  the  book. 

Practical  Applications  of  principles  to  business  should  be  intro- 
duced throughout  the  course,  thus  uniting  practice  with  theory,  and 
science  with  art,  and  making  the  pupil  both  a  scientific  scholar,  aijd 
a  iudicious  man  of  business. 

To  accomplish  this  would  of  course  require  years  of  labor,  care^;; 
and  study ;  but  the  pupil  thus  educated  would  finally  become  an 
honor  to  himself,  a   credit  to   his   teacher,    and   an   ornament. toj 
society.  -^^j^  jy^j^  laiAiiiOX 

Throughout  this  series  we  have  endeavored  to  follow  the  plan 
Jhere  laid  down,  with  what  success,  remains  yet  to  be  decided. 

But  whilst  a  work  may  be  prepared  with  care,  and  the  princi- 
ples presented  clearly  and  lucidly,  yet  the  instructor  must  sec- 
ond the  aim  of  the  author,  and  enforce  theories  and  the  method 
of  application  upon  the  attention  of  tfce  scholar,  and  be  sure  to 


XVlll  INTRODUCTORY    REMARKS. 

have  them  correctly  understood.     To  teach  with  success,  however, 
the  teacher  must  possess — 

1st.     A  thorough  knowledge  of  his  subject;  and 

2nd.     The  entire  covjidence  and  gojd  will  of  his  pu2nls. 

If  in  addition  to  this,  he  has  a  good  black-hoard  and  numerical 
frame,  he  need  not  fear,  but  push  on,  certain  of  success. 

The  first  book  of  these  series  will  propably  be  published  the  en- 
suing season.  It  is  recommended  that  this  or  some  other  mental 
work  should  be  studied  before  studying  written  Arithmetic. 

A  few  errors  were  discovered  in  this  work  after  the  forms  were 
printed,  and  are  mentioned  in  an  Errata  at  the  end  of  the  vol- 
ume. It  may  be  possible  some  errors  have  escaped  detection. 
Those  discovered  will  all  be  corrected  in  future  editions. 

Should  some  portions  of  this  work  be  thought  too  difficult  at 
first,  they  may  be  omitted,  until  a  review.  It  is  left  to  the  judgment 
of  the  teacher,  however,  what  part  to  omit,  and  what  to  study. 

In  teaching,  the  scholar  should  be  tanght  to  depend  upon  his  own 
resources.  Examples  should  be  explained  indirectly,  and  the 
method  adopted  by  some,  of  working  sums  for  scholars  on  the 
slate,  without  illustration,  should  be  discarded  at  once.  The 
teacher,  however,  should  not  permit  the  learner  to  pass  by  a 
question  or  principle  until  it  is  perfectly  familiar  to  him. 

In  conclusion,  the  writer  would  say  to  Teachers — Your  success 
must  depend  principally  upon  your  own  exertions.  You  have  an 
important  charge  committed  to  your  keeping,  and  therefore  yours  is 
a  heavy  responsibility.  Let  your  motto  then  be  "  Onward  and 
Upward.'*  And  whilst  you  instil  the  principles  of  science  into  the 
youthful  mind,  endeavor  to  train  it  in  such  a  manner  that  it  shall 
reflect  credit  upon  its  teacher,  and  honor  upon  itself.  Excite  in  the 
minds  of  your  pupils  a  noble  spirit  of  generous  rivalry  and  emula- 
tion ;  and  not  only  excite  this  spirit,  but  keep  up  the  excitement, 
and  your  success  will  astonish  even  yourself. 

To  Students  he  would  say — Persevere.  Never  give  up  a  question 
because  it  is  hard,  or  pass  by  a  demonstration  because  you  eannot 


XIX  INTRODUCTORY    REMARKS. 

comprehend  it  at  once.  Remember,  *' Where  there  is  a  will  there  is 
a  way,"  and  **  Nothing  so  hard  but  search  will  find  it  out.'*  If 
you  ever  expect  to  gain  a  knowledge  of  Arithmetic  without  dose 
application,  and  hard  study ,  you  will  be  sadly  disappointed.  There- 
fore meet  every  difficulty  ^'VVm^v/,  and  combat  every  obstacle  reso- 
liUely,  and  success  will  eventually  crown  your  efforts . 


f 

t 


llz:^ 


COMMON  ARITHMETIC. 


SECTION  I. 
NOTATION  AND  NUMERATION. 

A  single  thing  is  called  a  unit,  or  one  ;  one  and  one  more  arc 
called  two  ;  two  and  one  more  are  called  three  ;  three  and  one  more 
are  called  ybwr/  four  and  one  more  are  called  j'lw  /  five  and  one 
more  are  called  six,  (fee. 

The  expressions  one,  tvx),  three,  four,  &c.,  are  called  Numbers. — 
Hence: 

Observation  1.  Number  is  an  expression  signifying  a  unit,  era 
.collection  of  units.  It  always  answers  to  the  question,  *'How  ma- 
ny?'* Numbers  are  subject  to  certain  laws  and  regulations,  and  are 
/applied  to  practical  business.  These  laws  and  relations,  when  ar- 
ranged in  systematic  order,  form  what  is  called  an  Arithmetic. — 
Hence : 

Obs.  2.  Arithmetic  is  the  science  of  numbers.  It  is  both  theoreti- 
cal and  practical. 

Obs.  3.  The  theory  of  Arithmetic  consists  in  the  anahzation  of 
the  laws  and  principles  of  numbers. 

Obs.  4.  The  j5rac/tce  of  Arithmetic  consists  in  the  application  of 
its  principles  to  common  business  transactions. 

Remark. — Science  is  knowledge  reduced  to  order. 

Article  I.     Notation. 

Obs.  1.  Notation  is  the  method  of  expressing  numbers  by  char- 
acters or  signs. 

Obs.  2.  There  are  two  diflferent  methods  of  notation  in  use, 
called  the  Roman  and  Arabic  methods,  because  it  is  supposed  that 
these  nations  first  invented  them. 

What  is  a  single  thing  called?  What  does  unit  mean?  What  are  one  and 
more  called?  Two  and  one  more?  Three  and  one  more?  Four  and  ojia 
more?  Fire  and  one  more?  Six  and  one  more?  Seven  aid  one  more? 
Eight  and  one  more?  Nine  and  one  more?  What  are  the  expressions  onei 
'  two,  three,  &c.,  called?  What  then  is  number?  To  what  question  does  it 
answer?  To  what  are  numbers  subject?  To  what  are  they  applied?  What 
do  these  laws  and  regulations  form  when  arranged  in  order?  What  then  is 
Arithmetic?  How  is  Arithmetic  divided?  What  is  the  theory  of  Arithmetic? 
The  practice  of  Arithmetic?  What  is  Science?  What  is  Notation?  How 
many  methodi  of  Notttion  aro  thtr*  in  Uft?  What  are  tb«y7  Why  are  they 
eo  eailed? 

t 


COMMON    ARITHMETIC. 


Sect.  I 


Case  T.  The  Roman  Method. — By  the  Roman  method  we  ex- 
press numbers  by  the  use  of  the  seven  following  letters,  viz  :  I.  V. 
X.  L.  C.  D.  and  M. 

When  standing  alone,  their  values  are  as  follows  :  I.,  one  ;  V., 
five  ;  X.,  ten  ;  L.,  fifty ;  C,  one  hundred,  and  M.,  one  thousand  ; 
and  by  various  repetitions  and  combinations,  may  express  any  other 
number. 

Their  manner  of  use  is  clearly  shown  in  the  following 

TABLE. 

I  denotes ^ one 

n  "      two 

m  "       three 

IV  "       --_---_  four 

V  "       -  — five 

VI  "       six 

VII  " seven 

VIII  "       eight 

IX  "       nine 

X  "       -, ten 

XI  ** eleven 

XII  **       twelve 

Xni  «       - thirteen 

XIV  "       fourteen 

XV  "       fifteen 

XVI  "       sixteen 

XVII  "       seventeen 

XVIII  "       .  eighteen 

XIX  "       nineteen 

XX  "       twenty 

XXI  "       twenty-one 

XXII  "       twenty-two  (fee. 

XXX  "       thirty 

XL  «       forty 

L  "       fifty 

LX  *'       sixty 

LXX  "       --.   seventy 

LXXX         "       eighty 

XC  '* ninety 

C  **       one  hundred 

CC  **       - two  hundred 

CCC  "       __  three  hundred 

CCCC  "       _____ four  hundred 


How  do  we  express  numbers  by  the  Roman  method?  What  va'ue 
have  these  letters  when  standing  alone?  Can  any  number  be  expres»ed  by 
those  letters?    How? 


Art.  1.  NOTATION.  3 

D  denotes five  hundred 

DC  ** six  hundred 

DCC  **       seven  hundred 

DC  ''       eight  hundred 

DCCCC        "       nine  hundred 

M  "       one  thousand 

MDCCCXLVIII  denotes. _  _one  thousand,  eight  hundred  and 

forty-eight. 

By  examining  this  table,  it  will  be  perceived  that  when  a  letter  is 
repeated,  its  valzte  is  repeated;  as  II  two,  X.'K  twenty  (fee;  also,  when 
two  letters  of  different  values  are  joined  together,  if  the  ^655  be  placed 
before  the  greater,  the  greater  is  diminished  by  a  sum  equal  to  the 
value  of  the  less  ;  but  ii  the  less  be  placed  after  the  greater,  the 
value  of  the  greater  is  increased  by  a  sum  equal  to  the  value  of  the 
less.  Thus  :  X  denotes  ten  ;  but  IX  denotes  nine,  and  XI  denotes 
eleven. 

Obs.  3.     A  line  ( — )  placed  over  a  letter  increases  its  value  a 

thousand  times.  Thus  :  V  denotes  yit;e  ;  but  V  denotes  five  thou- 
sand. 

Note. — This  method  of  expressing  numbers  is  but  little  used,  except  in  de- 
noting Chapters,  Sections,  &.c. 

Case  2.  The  Arabic  Method. — Numbers  are  expressed  by  the 
common,  or  Arabic  method,  by  the  following  characters  or  figures, 
viz  : 

1,        2,         3,  4,         5,        6,  7,  8,         9,  0. 

one,    two,    three,    four,    five,    six,    seven,    eight,   nine,    cipher. 

Obs.  4.     'i  he  first  nine  characters   are  called  significant  figures, 
ujp  because  they  always  express  some  number.     They  are  also  termed 
J    digits,  from  the  Latin  word  digitus,  signifying^7?^er. 

r  Obs.  5.  The  last  character  (0),  when  standing  alone,  has  no 
value;  therefore  it  is  called  a  cipher,  which  mesLUs  nothing ;  but 
when  placed  at  the  right  of  a  significant  figure,  it  increases  the 
value  of  this  figure  ten  times, 

Numbers  larger  than  nine  are  expressed  by  different  combina- 
tions of  the  foregoing  characters.  Thus  :  ten  is  expressed  by  1 
..and  0,  thus,  10  ;  twenty  is  expressed  by  2  and  0,  thus,  20,  <fec. 


What  effect  does  it  have  upon  its  value  to  repeat  a  letter?  What 
effect  does  it  have  to  place  a  letter  expressing  a  smaller  number  before 
a  letter  expressing  a  larger  number?  To  place  the  smaller  after  the 
larger?  How  many  times  does  a  line  ( — )  placed  over  a  letter  in- 
crease its  value?  For  what  is  this  method  of  expressing  numbers  used? 
How  are  numbers  expressed  by  the  Common,  or  Arabic  method?  What  are 
the  names  of  these  characters?  What  are  the  first  nine  characters  called? 
Why?  What  other  name  are  they  termed?  From  what?  What  value  has 
the  character  (0)  when  standing  alone?  What  is  it  called?  What  does  cipher 
mean?  What  effect  does  it  have  to  place  a  cipher  at  the  right  of  a  significan 
figiire?    How  are  nurabers  larger  than  nia«  ©xpreseed ?  How  is  t«n  expreswd? 


COMMON    ARITHMETIC. 


Sect.  I. 


TABLE    OF   NUMBERS    FROM    TEN    TO    ONE    THOUSAND. 


ten 

eleven  _ . 
twelve  _- 
thirteen . 
fourteen, 
fifteei 


?n. 


sixteen 

seventeen . 
eighteen .  _ 
nineteen .  _ 
twenty 


twenty-one.  _ 
twenty-two_  _ 
twenty-three. 

thirty 

thirty-one 

forty 

fifty 

sixty .__. 

seventy 


eighty 

ninety 

one  hundred 

one  hundred  and  one. 
one  hundred  and  ten_. 
one  hundred  and  fifty. 

two  hundred 

three  hundred 

four  hundred 

five  hundred 

six  hundred 

seven  hundred m.. 

eight  hundred .__ 

nine  hundred 


is  expressed 10 

- H 

12 

13 

14 

15 

16 

17 

18 

19 

._- 20 

.. 21 

22 

23&C 

30 

31&C 

40 

50 

60 

70 

80 

_..     90 

100 

-_.    101 

___  no 

150 

200 

300 

400 

500 

600 

_. 700 

_   800 

900 


nine  hundred  and  ninety -nine  is  expressed 999 

one  thousand  is  expressed 1000 

Article  2.     Numeration- 

Obs.  1.  Numeraiion  is  the  art  of  expressing  numbers  hy  words y 
or  of  reading  numbers  expressed  hy  figures. 

In  Numeration,  diflferent  places  are  assigned  for  figures,  and  a 
name  given  to  these  places.     Hence — 


Twenty?    Thirty? 
Nttm«ritioii? 


Fifty?     One   hundred?     Fire   hundred?     What    is 


Art.  2.  NUMERATION.  B 

Obs.  2.     The  value  of  a  figure  depends  upon  the  place  U  occupies. 
Thus: 
4,  standing  alone,   expresses  simply  4  things,  or  4  units,  and 

is  written  thus .       4 

But  if  we  annex  a  cipher  to  4,  we  increase  its  value  ten  tiiries, 

and  make  it  ^  tens,  or /bn^y,  written  thus 40 

If  we  annex  two  ciphers  to  4,    we  increase  its  value  one  hun- 
dred times,  and  make  it  4  Imndred,  written  thus 400 

If  we  annex  three  ciphers  to  4,  we  increase  its  value  one  tlwu- 

sand  times,  and  make  it  \  thousand^   written  thus 4000 

Obs.  3.  Hence — Every  removal  of  a  figure  towards  the  left  in- 
creases its  value  ten  times  ;  or,  it  alioays  takes  10  units  of  any  07ie 
order  to  moke  1  of  the  next  higher  order. 

It  matters  not  whether  the  figures  at  the  left  are  significant  fig- 
ures or  ciphers,  the  increase  is  the  same.  Thus :  the  above  ex- 
pressions may  be  written  4444  ;  the  right  hand  figure  being  units, 
the  second  tens,  the  third  hundreds,  and  the  fourth  thousands  ;  and 
as  the  figure  in  each  order  is  4,  the  whole  expression  is  read/oz^r 
thousand,  four  hundred,  four  tens  (forty)  andyoe^r. 

The  different  places,  or  orders  for  figures,  and  their  names,  may 
be  exhibited  in  the  following 

TABLE. 


_r>  »J  M  c-< 

g  §  .2  § 


C3 


.  '5  .2       to       C.   o       -at 

.2  i2  02        Iq  CO  fl  M 


^73 


1457962057935428 

Note. — This  is  according  to  the  French  method.  The  English,  after  hun- 
dreds of  millions,  hundreds  of  billions,  hundreds  of  trillions,  &c.,  reckon 
thousands  of  millions,  tens  of  thousands  of  millions,  hundreds  of  thousands 
of  millions,  and  the  same  of  billions,  trillions,  &c.,  assigning  sLr  places  to  mil- 
lions, billions,  &c.,  instead  of  three. 

Upon  what  does  the  value  of  a  figure  depend?  What  does  the  num- 
ber 4,  standing  alone,  express?  If  we  annex  a  cipher  to  4,  what  does  it 
express?  If  we  annex  two  ciphers,  what  does  it  express?  Three  ciphers? 
What  effect  does  every  removal  of  a  figure  towards  the  left  have  upon  its 
value?  How  many  units  does  it  take  of  one  order  to  make  a  unit  of  the 
next  higher  order?  Does  it  make  any  difference  whether  the  figures  at  the 
iightare  significant  figures  or  ciphers?  How  la  liie  €xpie*jiioii  4444  read? 
Repeat  the  Table- 


'  4 


d  COMMON    ARITHMETIC.  SfeCt.  I. 

Obs.  4.     Every  number  must  ocaipy  one  or  more  places.     Thus: 

r  r  1 1 

The  number  s/.r  occupies  the  units  place  only  :  thus_^  -     -     -    6 
The  number  twenty-three  occupies  two  places,  viz  : 

units  and  tens ;  thus -      -      2    3 

The  number  two  hundred  and  sixty-seven  occupies 

three  places,  viz  :  iinits,  tens  and  hundreds  ;  thus-_    -     2     6    7 
The  number  two  thousand,  two  hundred  and  twen- 
ty-two occupies  four  places,  viz  :  units,  tens,  hun- 
dreds and  thousands  ;  thus 2     2     2    2 

From  the  example  we  perceive 

Obs.  5.  That  the  same  figure  has  different  values,  according  to 
the  place  it  occupies.  Hence,  we  infer  that  figures  have  two  values, 
viz  :  simple  and  local. 

Obs.  6.  The  simple  value  is  thevcdue  expressed  hy  the  figure  when 
standing  alone:  as  5  expresses  simply  five,  and  6  expresses  sim- 
ply six. 

The  local  value  is  the  value  expressed  hy  the  figure  when  com- 
bined vnth  other  figures,  and  depends  upon  the  p)lace  it  occupies. 
Thus  : 

In  the  number  55,  the  5  in  the  ?/««V'5  place  is  simply  5  ;  but  the  6 
in  the  ten's  place,  if  considered  with  reference  to  the  other  5,  be- 
comes 5  tens,  OY  fifty.  The  term  local  is  derived  from  the  Latin 
Avord  locus,  signifying  place. 

To  facilitate  the  reading  of  numbers,  the  orders  are  pointed  off 
into  periods  of  three  figures  each.  '1  he  first  right  hand  period  is 
called  the  unifs  period ;  the  second,  the  thousand's  period ;  the 
third,  the  million's  period,  &c.     1  his  may  be  illustrated  as  follows: 

CO 

§  §  i  i  . 

i§       :sg        s§        as 

O  .-^  o  r;2  o  .-;:;  o   c>     •» 

c^i!  r«j2  aifl  (/jISt!!  «J 

^g-c      ^%B      ^^SS      ^S-S      ^hS,^ 
Kr-H       ffi^pq       Kh^       Whh       WhP 

Trillion's         Billion's  Million's        Thousand's  Unit's 

period.  period.  period.  period.  period. 

Note. — The  English  have  six  places  in  each  period. 

How  many  places  will  the  number  six  occupy?     Twenty-three?     Two  hun- 
dred auj  sixty-seveu?     Two  thou8;iud,  two  hundredand  tw?nty-two?     Nam« 


Att.  2»  NUMERATION.  t 

As  we  have  said  before,  every  number  must  occupy  one  or  more 
places.     Thus :     The  number 

23  consists  of  2  tens  and  3  units,  and  is  called  twenty-three. 

365  consists  of  3  hundreds,  6  tens  and  5  units,  and  is  called  three 
hundred  and  sixty-five. 

1848  consists  of  1  thousand,  8  hundreds,  4  fpns  and  8  inits,  and 
reads  one  thousand,  eight  hundred  and  forty-eight.. 

How  many  units  and  tens  are  there  in  63? 

Ans.  3  units  and  6  tens. 

How  many  units  and  tens  are  there  in  47?  54?  27?  80?  36?  ^ 

How  many  units,  tens,  and  hundreds  are  there  in  126? 

Ans.  6  unitSy  2  tens,  and  1  hundred. 
How  many  units,  tens,  and  hundreds  are  there  in  216  ?  423?  108? 
240?  300?  789?  972? 

How  many  units,  tens,  hundreds,  and  thousands  are  there  in 
1827?  2362?  4786?  6721? 

How  many  units,  tens,  <fec.,  in  the  following  numbers  :  5?  18? 
20?  17?  60?  58?  40?  120?  111?  107?  209?  780?  2820?  2006? 
7460?  7000?  12067? 

Obs.  7.  To  numerate  figures — Commence  at  the  right  hand,  and 
call  the  orders  as  they  stand  in  the  Table. 

To  read  figures — Commence  at  the  right  Jiand,  and  point  them 
off  into  periods  of  three  figures  each ;  then  commence  at  the  left 
hand  and  read  each  period  as  if  it  stood  alone,  add.ng  the  name 
of  the  period. 

Numerate  and  read  the  number  623.  Ans.  3  u7hts^  2  tens,  6 
hundreds,  which  reads  six  hundred  and  twenty-three. 

In  like  manner  let  the  pupil  numerate  and  read  the  following 
numbers  : 


46 

126 

1026 

16932 

61728421 

78 

150 

1428 

214975 

74500112 

^Q 

197 

2764 

379681 

345678907 

47 

265 

4829 

127962 

763945316 

54 

390 

7008 

987943 

5438659123 

67 

411 

9060 

7237684 

29685734120 

75 

578 

10857 

4623794 

149769004721 

87 

976 

12764 

2776547 

691239468721623 

the  places  or  orders  occupied  by  each  of  these  numbers.  Can  the  same  figure 
have  different  values?  How?  fiow  many  values  have  figures?  What  e 
they?  What  IS  the  simple  value?  The  local  value?  Upon  what  does  the  lo- 
cal value  depend?  In  the  number  55,  what  is  the  value  of  the  5  in  the  unit's 
place?  Of  the  5  in  the  ten's  place,  when  considered  with  reference  to  the 
other  figure?  How  do  we  facilitate  the  reading  of  numbers?  How  many 
figures  does  each  period  embrace?  W^hat  is  the  first  period  called?  The  sec- 
ond? The  third?  The  fourth?  The  fifth?  How  many  places  does  ...e 
nuiiibir  23  occupy?  365?  1848?  What  are  the  names  of  the  places  e  ch  of 
these  figures  occupy?  How  do  we  numerate  figures?  Where  do  we  com- 
mence to  point  off  figures  into  periods?  Where  do  we  comnitnce  to  rMd? 
How  d©  we  read  each  p«ri.©d? 


COMMOK    ARrrnMBTIO. 


Sect.  !• 


The  number  forty -six  consists  of  6  units  and  4  tens  ;  there- 
fore to  write  forty-six,  we  place  a  6  in  the  unifs  place,  and 
a  4  in  the  ^ew'*  place,  thus _ 46 

The  number  two  hundred  and  six  consists  of  2  hundreds, 
0  tens,  and  6  units,  and  is  written  thus 206 

The  number  one  thousand  and  sixty  consists  of  1  thousand,  0 
hundred i  6  tens,  andO  units,  and  is  written  thus . 1060 

NoTK. — When  unit*  only  are  mentioned,  the  right  band  place  is  always  un- 
derstood. 

Hence,  to  write  numbers,  we  have  this 

Rule. —  WrUe  each   significant  figure  in  the  order  which  it  be- 
longs, and  place  a  cipher  in  all  vacant  orders. 

EXERCISES    FOR    THE    SLATE. 

Should  the  learner  find  any  diflficulty  in  reading  the  following 
numbers,  it  might  assist  him  to  proceed  thus : 

Draw  a  num- 
ber of  lines,  and 
write  the  names 
of  the  orders  in 
their  respective 
spaces ;  then  by 
distinguishing 
the  periods  by 
double  lines,  the 
numbers  can  be 

easily  written  and  read.  Thus  :  to  write  three  hundred  and  forty 
six  millions,  one  hundred  and  four  thousand,  and  twenty-seven,  we 
write  each  figure  in  its  order,  and  place  a  cipher  in  the  orders  left 
vacant,  and  then  it  is  easily  read,  as  above. 

1 2  Five  hundred  and  ninety-nine. 

13  Seven  hundred. 

14  Eight  hundred  and  three. 

15  Nine  hundred  and   fifty  one. 

16  One  thousand  and  four. 

17  Three    thousand,   four  hun- 
dred. 

18  Six  thousand,  four  hundred 
and  thirty. 

19  Nine   thousand,    three    hun- 
dred and  twenty- six. 

20  Ten  thousand,  one  hundred. 


o 

» 

03 

■< 

-d 

^ 

i 

1 

^ 

^ 

h) 
^ 

Ti 

H 

-TIJ 

d 

1 

^ 

1 

C 

1 

> 

3 

4 

6 

1 

0 

4 

0 

2 

7 

Write  twenty-five. 

Forty-eight. 

Fifty-four. 

Seventy-six. 

Eighty-nine. 

Ninety-seven. 

One  hundred  and  seven. 

One  hundred  and  thirty-two. 

Two  hundred  and  forty. 

10  Three    hundred    and    sixty - 
five. 

1 1  Four  hundred  and  eighty-six. 


How  do  we  write  the  number  forty-six?  Two  hundred  and  six?  One  thou- 
•and  and  sixty?  When  units  only  are  mentioned,  what  place  i«  understood? 
What  is  the  rule  for  writing  numbers? 


SIMPLE   ADDITION. 


0 


21  Twenty-two     tliousand     and!  28 
seventy-three. 

22  Sixty-four  thousand,  nine 
hundred  and  seventy-seven . 

S3  Eighty  thousand,  three  hun- 
dred and  one. 

24  One  hundred  and  four  thou- 
sand and  eight. 

25.  Seven  hundred  thousand, 
nine  hundred, 

S6.  Nine  hundred  thousand,  two 

hundrod  and  forty-seven.  one 

27.  One  million,  sixty  thousand 
and  nine. 

Note. — The  method  o£  reading  and  writing  numbers  should  be  clearly  ex- 
plained on  the  black-board  by  the  teacher,  ontil  the  schclar  thorougklg  under- 
stands il,  and  can  read  or  write  any  number  readily.     A  little  laber  a«d  exer- 
tion  on  the  part  of  the  teacher  will  be  found  to  be  mere  beneficial  and  satis- 
Ci/L^T factory  than  an  elaborate  treatise  on  the  subject. 


Three  millions,  two  hundred 
atid  seventy-six  thousand. 

29  Eight  hundred  and  forty  mil- 
lions. 

30  Eight  billions,  ninety-seven 
millions,  six  thousand,  seven 
hundred  and  nine. 

31  Six  hundred  and  ninety-seven 
trillions,  two  hundred  and  four 
billions,  seventy  millions,  four 
hundred  and  six  thousand  and 


SECTION  IL 
r-  SIMPLE  ADDITION. 

Artrxe  1.     Mentai.  Exercises, 

1 .  Charles  bought  a  book  for  8  cents,  a  slate  for  6  cents,  mA  a 
pencil  for  1  cent.     How  maaay  cents  did  he  pay  for  the  whole  ? 

Ans.   15  coats, 
8  cents,  and  6  cents,  and  X  cent  are  how  many  cents  ? 

2.  John  gave  6  cents  for  a  writing-book,  6  cents  for  an  ink-staxid, 
and  4  cents  for  some  quills.     How  many  cents  did  he  give  for  all^ 

6  and  6  and  4  are  how  many? 

3.  A  poor  boy  met  some  ladies,  one  of  whom  gave  him  3  cents, 
smother  6  cents,  anotiver  7  cents,  and  anot^ier  9  cents.  How  many 
-cents  did  he  get? 

4.  A  man  paid  8  dollars  for  some  sheep,  7  d-ollars  for  some  .K(i»gs, 
and  2  dollars  for  some  fowls.     How  many  dollars  did  he  spend? 

5.  A  lady  bought  4  yards  of  cloth  in  one  piece,  Q  in  another,  5 
in  another,  7  in  another,  and  9  in  another.  How  many  yards  did 
«he  buy? 

6.  John  has  8  marbles,  William  6,  Thomas  9,  Henry  5,  Charles 
7,  James  3,  and  Robert  4.     How  many  have  they  all? 

7.  One  boy  has  6  apples,  another  7  apples,  another  4  apples, 
another  8  apples,  and  another^  apples.     How  many  have  they  all? 

8.  One  man  has  6  horses,  another  4,  another  9,  another  3,  aaa©- 
ther  8,  and  another  7.     How  many  have  they  all? 

a2 


la 


COMMON    ARITHMETIC. 


Sect.  II 


9.  On  one  slieir  are  9  books,  oa  anoiher  7,  on  another  8,  on  ano- 
ther 6,  and. on  another  5.     How  many  are  there  in  all? 

10.  I  bought  a  book  for  4  shillings,  a  slate  for  2  shillings,  some 
paper  for  3  shillings,  and  a  map  for  9  shillings.  How  many  shil- 
lings did  I  spend? 

1 1 .  A  man  sold  some  wheat  for  8  dollars,  some  corn  for  5  dol- 
lars, and  some  oats  for  7  dollars.     How  much  money  did  he  get? 

12.  4,  and  7,  and  5,  and  3,  and  9,  and  2,  and  1,  and  6,  and  8, 
are  how  many? 

Article  2.     Definitions. 

Obs.  1.  The  putting  together  two  or  more  numbers  to  find  one 
number,  as  in  the  preceding  examples,  is  called  Addition  ;  and  the 
number  thus  obtained  is  called  the  Sum.* 

Addition  may  be  of  Simple  or  of  Compound  numbers. 

Obs.  2.  Simple  Addition  is  when  the  numbers  all  express 
things  of  the  same  name  or  hind  ;  as  all  dollars,  all  yards,  (fee. 

Obs.  3.  Signs. — A  cross  (-}-)>  o"^  ^^^^  horizontal  and  the  other 
perpendicular,  is  thd  sign  of  addition.  It  is  called  7:>/ws,  which  is  a 
Latin  word,  meaning  more.  It  signifies,  that  the  numbers  between 
which  it  stands  are  to  be  added  together. 

Two  parallel  horizontal  lines  (=)  are  the  sign  of  equality.  It 
shows  that  the  number  before  it  is  equal  to  the  number  after  it. 
Thus:  6-f-4=10. 

To  facihtate  the  addition  of  numbers,  we  subjoin  the  following 
ADDITION  TABLE. 


2  aiul     I 
1  ae    3 
y    "      4 

3  "      5 

4  ♦       G 

5  "       7 

6  "      8 

7  "      9 

8  "     10 

9  "  n 

10  •'    12 


3  and 

1  are 

2  '• 

3  " 

4  " 


4  and 

5ai 

1(1 

4 

1  are    5 

1  ar 

'.    G 

5 

2    "      6 

2     " 

; 

6 

3    "      7 

3     " 

8 

7 

4    '•      8 

4    " 

9 

8 

5    «'      9 

5     '' 

10 

J> 

0    "    10 

()     '• 

U 

V-) 

7    "    11 

7    ' 

u 

11 

8     "    1-2 

8    " 

1:5 

12 

9     "    13 

9    *' 

14 

13 

10  '•    14 

10  " 

15 

1  a^e 


1  a  e    9 

2  •'     10 


1  are  10  I 


()   ' 

11 

7    ' 

r> 

8    • 

•    16 

9    ' 

17 

10  ' 

18 

13  ! 

14  i 

15  1 

Si 


The  pupil  should  be  frequently  exercised  in  mental  addi- 
tion, un  il  he  \'s>  perfectly  familiar  with  the  operation.  Care  should 
be  taken,  however,  that  the  questions  are  not  answered  mechanic- 
ally, from  a  knowledge  of  the  regular  increase  of  numbers.  This 
may  be  prevented  by  asking  promiscuous  questions. 

7  -|-  6  =  how  many  ? 

9  -|-  7  ==  how  many  ? 

9  -j-  4  =  how  many  ? 

4  -j-  3  -|-  5  ==  how  many?  - 

What  is  Addition?  Wliat  isthesnm?  How  may  Addition  be  divided?  What 
is  Simple  Addition?     What  is  the  sign  of  Addition''     What  does  pins  mean? 


*By  many  Authors,  tlje  result  obtained  by  adding  two  or  more  numbers  lo;:cther,  is  called 
the  Amount:  but  the  term  .>?//>£>««*,  we  think  properly  beionga  to /n«ffr«/,  aiid  ahoukl  not 
be  applied  to  Addition. 


Art.  2.  SIMPLE  ArmrioN.  11 

ft 

9  _j_  4  _j_  7  —  ho-vr  many? 

6  "4-  3  -j-  4  — -  how  many? 

6-4-24-7  +  8  =  how  many? 

7+8-[-5-|-2  =  how  many  ? 
"'-"".       6  +  8  4-  2  +  4  +  1  =  how  many? 
,.:  7  +  2  +  1  -|-  4  -|-  3  =  how  many? 

3  -|-  2  -|-  1  -!-  9  -|-  7  =  how  many? 

9  _|_  7  -|_  5  -|-  6  ^|-  2  -|-  1  -I-  3  =  how  many  ? 

Example  1.  A  man  has  2  orchards;  one  has  24  trees,  and  the 
other  has  32  trees.     How  many  trees  in  both  orchards? 

SoliUlon. — In  24  are  2  tens  and  4  imUs ;  in  32  are  3  tens  and  2 

units.     Now  units  and  tens  cannot  be  added  together,  because  it 

takes  ten  units  to  make  one  ten  ;  but  we  can  add  units  to  units,  and 

f-.  ,  tens  to  tens  ;  and  4  units  and  2  units  are  6  units,  and  2  tens  and  3 

tens  are  5  tens,  and  5  tens  and  6  units  are  56.  Ans.  6Q. 

Or  we  may  set  the  numbers  down  on  the  slate,  and  32 
add  them,  recollecting  to  place  units  under  units,  and  24 
tens  under  tens.     Thus  :  — 

bQ 

Obs.  4.  We  place  units  under  units,  tens  under  tens,  <^c.,  because 
it  is  more  convenient  to  add  them  when  placed  in  this  manner,  as  we 
can  only  add  together  numbers  of  the  same  name.     Hence — 

To  add  numbers : 

Obs.  5.  Write  units  under  units,  tens  under  tens,  hundreds  under 
hundreds,  d'c. 

Commence  at  the  right  hand.  Add  each  column  separately,  and 
place  the  result  beneath  it. 

Obs.  6.  Proof. — Begin  at  the  top  and  add  downwards  ;  if  the 
two  results  are  alike,  the  work  is  correct. 

1 .  We  add  the  numbers  from  the  top  downwards,  in  order  to  take 
them  in  a  different  manner,  it  being  hardly  probable  that  the  same 
mistake  would  occur  by  both  processes,  unless  it  was  intentional. 

2.  The  learner  must  not  think  when  he  takes  up  the  slate,  that 
he  can  dispense  with  thinking  and  reasoning.  This  is  not  the  case. 
He  should  exercise  his  mental  faculties  as  much  when  he  uses  the  slate 
as  ivhen  he  does  not.  His  slate  is  merely  used  for  convenience  in  set- 
ting down  the  operation  when  it  is  too  long,  or  the  numbers  too  large 
to  be  easily  retained  in  the  mind.     That  is,  it  is  used  to  assist,  and 

^  not  to  take  the  place  of  the  mental  faculties. 

mk        What  does  it  signify?  What  is  the  sign   of  equality?     What  does  it  show? 

?*r  How  do  we  set  the  numbers  down  to  add?  Why  do  we  place  units  under 
units,  tens  under  tens,  &.C.?  Where  do  we  commence  to  add?  Where  do  we 
place  the  sum  of  each  column?  How  do  we  prove  the  operation?  Why  do 
we  add  the  numbers  from  the  top  downwards?  Should  the  scholar  dispense 
with  thinking  and  reasoning  when  he  takes  up  the  slate?  What  is  the  use  of 
the  Elate? 


IS  COMMON    ARtTHMETIC.  ScCt,  II. 

2.  In  one  field  are  127  sheep,  in  another  231,  and  in  another 
341.     How  many  are  there  in  all  the  fields?  Ans.  699. 

3.  In  one  school  are  62  scholars,  in  another  123,  and  in  another 
114.     How  many  are  there  in  all?  Ans.  299. 

4.  One  man  has  324  dollars,  another  232  dollars,  another  2123 
dollars,  and  another  6210  dollars.  How  many  dollars  have  they 
all?  Ans.  8889. 

5.  In  one  book  are  240  pages,  in  another  320  pages,  m  another 
1112 'pages,  in  another  1023  pages,  and  in  another  1304  pages. 
How  many  pages  in  all?  Ans.  3999. 

6.  241  -|-  122  -[-  1234  ~|-  21201  -j-  45101  =  how  many? 

Ans.  67899. 

7.  1234 -1-20412 -[-30211  -|- 12132 -|- 24000  =  how  many? 

Ans.  87989. 

Articlb  3.     Carrying  in  Addition. 

Ex  1.  A  man  has  76  bushels  of  grain  in  one  box,  and  58  in 
another  box.     How  many  bushels  has  he  in  both? 

SoliUion. — Here  a  difficulty  presents  itself,  as  8  units  -|-  6  units 
=  14  units,  which  cannot  be  expressed  by  one  figure;  but  the 
learner  must  recollect  that  14  units  =  1  ten  and  4  units ;  we  will 
therefore  write  the  4  units  under  the  units,  and  reserve  the  1  ten  for 
the  next  column.  Thus:  8  units -|- 6  units  =  14  units,  which  is  1  ten 
and  4  units  ;  5  tens  -[-  7  tens  =12  tens,  -|-  1  ten,  which  was  re- 
served =  13  tens,  which  is  1  hundred,  and  3  tens ;  and  the  whole 
result  is,  1  hundred,  3  tens,  and  4  units,  or  134.  Ans.  134. 

Hence — ^When  the  sum  of  any  column  exceeds  9 — 

Obs.  1 .  Set  down  the  unit  figure  only,  and  carry  the  tens  to  the 
next  column  ;  or  in  other  ivords,  carry  1  for  every  10. 

This  principle  may  be  illustrated  as  follows : 

2.  Let  it  be  required  to  add  together  446  and  678.  uiME' 

The  sum  of  6  units  -|-  8      Operation.  jBHE 

units  =   14  units,  or  1  ten         678  ^^^ 

and  4  units,  which  we  set         446 

down   accordingly  :    4   tens         

-|-  7  tens  =11  tens,  or  1 
hundred  and  1  ten,  which 
we  write  in  their  respective 

orders  :    4  hundreds  and   6       

hundreds  =    10  hundreds,       1124  =  the  total  sum. 
or  1  thousand,  0  hundreds,   which  we  also  place  in  their  proper 
places ;  then  adding  the  several  sums  together,  as  they  stand,  we 
obtain  1 124  as  the  total  sum  or  answer.  Ans.  1 1 24. 

How  do  w«  proceed  when  the  lum  of  any  order  exccedi  9?    Explain  the 
priaoiplo  of  currying. 


14  = 

:  the  sum  of  the  units. 

11  = 

"     tens. 

0  = 

"       "     hundreds. 

Art.  3.  SIMPLE   ADDITION.  13 

3.  Add  together  2467,  1924,  2761,  and  6279,  and  prove  the  ope- 
ration. 

2467 
1924 
2761 
6279 

Sum      13431         As  the  two  results  are  alike,  the  work  is  sup- 

posed  to  be  correct. 

Proof.   13431 

From  the  preceding  illustration,  we  derive  the  following 

GENERAL  RULE  FOR  ADDITION. 

I.  IVriie  the  numbers  to  be  added  so  that  the  figures  of  the  same 
order  may  stand  directly  under  each  other,  and  draw  a  line  beneath. 
(Art.  2.  Obs.  5.) 

II.  Begin  at  the  right  hand  and  add  each  column  separately  ;  and 
if  the  sum  be  9,  or  less,  write  it  under  the  column  added.  (Art.  2. 
Obs.  5.) 

III.  But  if  the  result  should  exceed  9,  set  doivn  the  unit  figure  ,of 
the  result,  and  carry  the  tens  to  the  next  column.     (Art.  3.  Obs.  1.) 

IV.  Set  down  the  ivhole  sum  of  the  left  hand  column. 

Proof. — Begin  at  the  top  and  add  downwards  ;  if  the  two  results 
are  alike,  the  work  is  correct.     (Art.  2.  Obs.  6.) 

EXERCISES    FOR   THE    SLATE. 

1.  A  man  bought  a  horse  for  85  dollars,  and  a  wagon  for  67 
dollars.     What  did  they  both  cost  him?  Ans.   152  dollars. 

2.  A  man  has  64  sheep  in  one  field,  and  148  in  another  field. 
How  many  sheep  has  he?  Ans.  212. 

3.  A  man  being  asked  his  age,  answered  :  **  I  was  $7  years  of 
age  when  I  married,  and  have  been  married  48  years."  How  old 
was  he?  Ans.  75  years. 

4.  A  man  has  owing  to  him — from  A.  463  dollars,  from  B.  798 
dollars,  from  C.  840  dollars,  from  D.  56  dollars,  and  from  E.  15 
dollars.     How  much  has  he  owing  him?  Ans.  2172  dollars. 

5.  A  man  has  4t3  bushels  of  wheat,  291  bushels  of  oats,  1479 
bushels  of  corn,  and  276  bushels  of  barley.  How  many  bushels 
of  grain  has  he  in  all?  Ans.  2469. 

6.  A  man  received  1243  dollars  for  horses,  97f  dollars  for  cat- 
tle, 672  dollars  for  grain,  427  dollars  for  sheep,  197  dollars  for 
hogs,  and  has  327  dollars  at  home.  How  many  dollars  has  he  in 
all?  Ans.  3838. 

7.  A  man  has  4  children :  one  is  1 8  years  old,  another  is  22 
years  old,  another  is  16  years  old,  and  the  other  is  12  years  old; 

What  if  the  g«n«r«l  rult  for  Additi«nt    Tht  proof? 


14  COMMON    ARITHMETIC.  Scct.  IL 

the  father's  age  is  22  years  more  than  the  '-'•"^-^   '^*"  hAr,  children's 
ages,     Required — the  age  of  the  father.  80  years, 

8.  A.  owes  one  man  642  dollars,  another  927  aoiiars,  another 
745  dollars,  and  another  2457  dollars.  What  is  the  sum  of  his 
debts?  Ans.  4771  dollars. 

9.  A  merchant  commenced  trading  with  14768  dollars  ;  the  first 
year  he  gained  1794  dollars,  the  second  year  he  gained  2986  dol- 
lars, the  third  year  he  gained  3789  dollars,  and  the  fourth  year  he 
gained  4697  dollars.     How  much  had  he  then^ 

Ans.  28034  dollars. 

10.  A  land-holder  has  in  one  farm  1217  acres,  in  another  976 
acres,  in  another  840  acres,  and  in  another  679  acres.  How  many 
acres  has  he?  Ans.  3712. 

11.  A  boy  walked — one  day  8  miles,  another  day  12  miles,  ano- 
ther day  17  miles,  and  another  day  23  miles.  How  many  miles 
had  he  walked  in  all?  Ans.  60. 

12.  A  drover  has  in  one  field  329  sheep,  and  142  lambs  ;  in 
another  field  he  has  476  sheep,  and  216  lambs;  and  in  a  third  field 
he  has  512  sheep,  and  319  lambs.  How  many  sheep  has  he? — 
How  many  lambs?     How  many  sheep  and  lambs  together? 

Ans.  to  the  last.   1994. 

13.  Four  men  traded  in  partnership.  A.  put  in  1269  dollars,  B. 
put  in  7642  dolhirs,  C.  put  in  3768  dollars,  and  D.  put  in  5429  dol- 
lars.    How  much  did  they  all  put  in?  Ans.   18108  dollars. 

14.  A  gentleman  owns  a  farm  worth  8478  dollars,  a  store  worth 
12694  dollars,  a  house  and  lot  worth  8621  dollars,  a  vessel  worth 
10216  dollars,  and  has  14376  dollars  worth  of  other  property. — 
How  much  is  he  worth  in  all?  Ans.  54385  dollars. 

15.  A  man  left  his  property  to  his  wife,  two  sons,  and  three 
daughters  :  to  his  Avife  he  gave  14612  dollars,  to  each  of  his  sons 
he  ave  8279  dollars,  and  to  each  of  his  daughters  he  gave  6297 
dollars.     What  was  the  value  of  his  property? 

Ans.  50061  dollars. 

16.  A  man  has  paid  on  a,  note — at  one  time  417  dollars,  at  ano- 
ther time  312  dollars,  at  another  time  512  dollars,  and  at  another 
time  794  dollars  ;  he  now  has  238  dollars  to  pay.  How  much  was 
the  note?  Ans.  2273  dollars. 

17.  The  distances  on  the  Ohio  canal  are  as  follows  :  from  Ports- 
mouth to  Chillicothe  51  miles  ;  from  Chillicothe  to  Circleville  22 
miles  ;  from  Circleville  to  Newark  60  miles  ;  from  Newark  to  Ros- 
coe  41  miles  ;  from  Roscoe  to  Dover  42  miles  ;  from  Dover  to  Mas- 
silon  28  miles  ;  from  Massilon  to  Akron  27  miles  ;  from  Akron  to 
Cleveland  38  miles.     Required — the  length  of  the  Ohio  canal. 

Ans.  309  miles. 

18.  A  man  bought  a  span  of  horses  for  312  dollars,  a  carriage 
for  497  dollars,  and  a  harness  for  76  dollars.  What  did  they  all 
cost?  Ans.  884  dollars. 


f 


Art.  3.  SIMPLE   ADDITION.  15 

19.  A  merchant  bought  calico  to  the  amoTint  of  649  dollar?. 
bro.idcloth  to  the  amount  of  837  dollars,  and  other  goods  to  \liv 
amount  of  497  dollars.     What  did  they  all  cost? 

Ans.   1983  dollars. 
20  The  above  merchant  sold  his  goods  so  as  to  gain  178  dollars 
on  the  calico,  279  dollars  on  the  broadcloth,   and  125  dollars  on 
the  other  goods.     How  much  did  he  gain,  and  for  what  sum  did  he 
sell  the  lot?  .  ^      ^    He  gained  582  dollars. 

^     (    He  sold  the  lot  for  2565  dollars. 

21.  There  are  four  numbers:  the  first  is  1248,  the  second  is 
2397,  the  third  is  as  much  as  the  first  and  second,  and  the  fourth  is 
as  much  as  the  second  and  third.  Required — the  third  and  the 
fourth  numbers,  and  the  sum  of  the  four. 

Ans.  to  the  last.   13332. 

22.  After  a  battle  it  was  found  that  7620  men  were  killed,  9276 
were  wounded,  792  had  deserted,  5874  had  been  taken  prisoners, 
1892  were  missing,  and  32716  were  left  fit  for  action.  Of  how 
many  men  did  the  army  consist  at  first?  Ans.  58170. 

23.  Tlie  second  Punic  war  commenced  529  years  after  the  found- 
ing of  Rome  ;  Carthage  was  destroyed  78  years  later  ;  the  Chris- 
tian Era  commenced  146  after  this  ;  the  death  of  Nero  was  69 
years  later  ;  242  years  afterwards  Constantine  ascended  the  Roman 
throne,  and  the  dissolution  of  the  Roman  Empire  occurred  about 
165  years  after.     Required — the  age  of  the  Roman  Empire. 

!  Ans.   1229  years. 

24.  Find  the  sum  of  the  following  numbers :  two  hundred  and 
four  ;  seven  hundred  and  thirty-nine ;  one  thousand  and  seventeen  ; 
three  thousand,  seven  hundred  and  sixty  ;  fifteen  thousand,  nine 
hundred  and  nine  ;  three  hundred  and  six  thousand,  one  hundred 
and  eight.  Ans.  337737. 

25.  Add  the  following  numbers  :  two  thousand  and  one  ;  seven 
thousand,  four  hundred  and  seventy-nine ;  eleven  thousand,  one 
hundred  and  eleven  ;  two  hundred  and  twelve  thousand  and  ninety- 
five  ;  eight  hundred  thousand  and  one ;  nine  hundred  and  ninety- 
nine  thousand,  nine  hundred  and  ninety-nine.         Ans.  2032686. 

26.  In  one  book  are  two  hundred  and  thirty-two  pages,  in  ano- 
ther are  two  hundred  and  sixty-four  pages,  in  another  are  three 
hundred  and  forty-six  pages,  and  in  another  are  two  hundred  and 
ninety  five  pages.     How  many  pages  in  all?  Ans.   1 137. 

27.  A  man's  farm  cost  him  eighteen  thousand,  five  hundred  dol- 
lars ;  a  store  cost  him  twenty  thousand,  two  hundred  and  fifty  dol- 
lars ;  a  house  and  lot  ten  thousand  and  fifty  dollars,  and  a  boat 
twelve  thousand,  five  hundred  dollars.  What  was  the  cost  of  the 
whole?  Ans.  61300. 

28.  Required— the  sum  of  12345  -\-  67890  ~|-  96432  -!-  7456 
-I- 6217 -i- 149321-!- 819360.  Ans.   1159021. 


16  COMMON    ARITHMETIC.  Se«t»  IIL 

29.  Required— the  sum  of  97452 -|-  68714-|- 127983-|- 15791 
--6829467  -|-  89'32164-|-  187437621  ~j-  19734653  -j-  890072 
-- 6214573  -|- 9876543210.  Ans.   10106891700. 

30.  In  1840  the  New  England  States  contained  2234822  inhabit- 
ants, the  Middle  States  contained  4604345,  the  Southern  States 
contained  5067843,  the  Western  States  contained  4984097,  the 
Territories  contained  128534,  the  District  of  Columbia  contained 
43712,  and  on  board  vessels  of  war  were  6100.  Required — the 
population  and  Naval  service  of  the  United  States  in  1840. 

Ans.  17069453. 


SECTION  III. 
SIMPLE  SUBTRACTION. 

Article  1.     Mental  Exercises. 

1 .  John  had  8  apples,   and  gave  3  of  them  to  his  sister.     How 
many  had  he  left? 

Take  3  from  8,  and  how  many  remains?  Ans.  5. 

2.  Mary  had  12  pins,  and  lost  5  of  them.     How  many  had  she 
left? 

3.  A  boy  having  18  cents,  bought  a  slate  for   10  cents.     How 
many  cents  had  he  remaining? 

4.  A  boy  had  9  marbles,  and  gave  his  brother  5.     How  many 
had  he  left? 

5.  A  lady  having  16  yards  of  cloth,  cut  ofif  9  yards.     How  many 
yards  were  left? 

6.  A  man  having  14  dollars,  paid  away  7  dollars.     How  many 
dollars  had  he  then? 

7.  James  went   15  miles,   and  John  5.     How  much  farther  did 
James  go  than  John? 

8.  Henry  had  12  cents,   and  spent  6  of  them.     How  many  had 
he  left? 

9.  William   made    17  marks   on  his  slate,  and  rubbed  out  8  of 
them.     How  many  marks  were  there  left? 

10.  John's  book  had  20  pages  in  it,  and  he  tore  out  10  of  them. 
How  many  pages  were  left? 

11.  Fourteen  boys  were  standing  together,  and  5  of  them  went 
away.     How  many  remained? 

12.  A  man  having  22  sheep,  sold  11  of  them.     How  many  had 
he  left? 

Article  2.     Definitions,  &c. 
Obs.  1       The  finding  the  difference  between  two  numbers,  as  in 
the  preceding  examples,  is  called  Subtraction.     It  may  be  defined 
— the  talcir\g  qf  a  Uis  number  from  a  gre&itr. 


Art.  2. 


SIMPLE    SUBTRACTION. 


It 


Subtraction  may  be  either  of  Simple  or  of  Compound  numbers. 

Obs.  2.  Simple  Subtraction  is  when  the  numbers  all  express 
things  of  the  sam£  nmm,  or  hind — as  all  dollars,  all  cents,  all 
yards,  &c. 

Obs.  3.  The  greater  number  is  called  the  Minuend,  (which  means 
to  he  diminished.)  The  lesser  number  is  called  the  Subtrahend, 
(which  means  to  be  svhtracted .)  The  result,  or  answer,  is  called 
the  Difference,  or  Remainder. 

Obs.  4.  Sign. — The  sign  of  Subtraction  is  a  short  horizontal 
line  ( — ),  called  Minus,  which  means  less,  and  signifies  that  the 
number  a/ter  it  is  to  be  taken  from  the  number  before  it.  Thus  : 
8 — 5=3  shows  that  5  taken  from  8  leaves  3,  and  reads,  8  minus  5 
is  equ^il  to  3. 

To  facilitate  the  progress  of  the  learner,  we  subjoin  the  following 


SUBTRACTION  TABLE. 

1  from 

2  from 

3  from 

4  from 

5  from 

1  leaves  0 

2  leaves  0 

3  leaves  0 

4  leaves  0 

5  leaves  0 

2      "     1 

3      *'      1 

4      "      1 

5      '*      1 

6       **     1 

3      ^'     2 

4      "     2 

5      "     2 

6-^2 

7       "    2 

4      "     3 

5      "     3 

6      "     3 

7      **     3 

8       •'     3 

5      "     4 

6      "     4 

7       «     4 

8      **     4 

9       "    4 

6      '*     5 

7      "     5 

8      ''     5 

9      ''     5 

10     "    5 

7      "     6 

8      «     6 

9      '^     6 

10    "     6 

11     "    6 

8      "     7 

9      "     7 

10    *'     7 

11     '^     7 

12     "     7 

9      "     8 

10    "     8 

11     *'     8 

12    "     8 

13     "     8 

10    *'     9 

11    ''     9 

12    "     9 

13    "     9 

14     "    9 

6  from 

7  from 

8  from 

9  from 

10  from 

6  leaves  0 

7  leaves  0 

8  leaves  0 

9  leaves  0 

lOle'vesO 

7      "     1 

8      "      1 

9      "      1 

10    "     1 

11     "     1 

8      "     2 

9      "     2 

10    "     2 

11     "     2 

12     "    2 

9      *«     3 

10    «     3 

11    "     3 

12    "      3 

13     "    3 

10    "     4 

11     "     4 

12    "     4 

13    '*     4 

14     *'     4 

11     "     5 

12    *'     5 

13    "     5 

14    "     5 

15     *'     5 

12    "     6 

13    ''     6 

14    "     6 

15    '*     6 

16     *'    6 

13    "     7 

14    "     7 

15    *'     7 

16     "     7 

17     "     7 

14    "     8 

15    "     8 

16    ''     8 

17    "     8 

18     "     8 

15    "     9 

16    "     9 

17    «     9 

18    "     9 

19     "     9 

Note. — This  Table  should  be  thoroughly  committed  to  memory  before  pro- 
ceeding farther. 


What  is  Subtraction?  How  may  it  be  defined.?  How  divided?  What  is 
Simple  Subtraction?  What  is  the  greater  number  called?  What  does  min- 
uend mean?  What  is  the  lesser  number  called  ?  What  does  subtrahend 
mean?  What  is  the  result,  or  answer,  called?  What  is  the  sign  of  Subtrac- 
tion? What  does  minus  mean?  What  does  it  signify?  What  does  8 — 5=3 
■how?    How  is  it  read? 


19 


COMMON    ARITHMETIC. 


Sect.  III. 


6 
14 

17 
12 


4  ==  how  many? 
3  =  how  many? 
7  =  how  many? 

5  =  how  many? 


18  —  10  =  how  many? 


17  —  6  =  how 


many 


18- 

-7- 

17- 

-  8- 

14- 

-  5- 

12- 

-3- 

16- 

-  4- 

1-9 


3  =  how  many? 
9  =  how  many? 
6  =  how  many? 
9  =  how, many? 
8  =  how  many? 


10  -|-  7  =  how  many? 

Obs.  5.  A  line,  or  vinculum  ( — ),  drawn  over  two  or  more  num- 
bers, signifies  that  they  are  to  be  taken  together  as  one  number.  A  pa- 
renthesis (  )  is  also  sometimes  used  for  the  same  purpose. 


Thus 


18  —.  4  -|-  2,  or  18  —  (4  -|-  2)  signifies  that  4  -I-  2,  or  6,  is  to  be 
taken  from  18,  which  leaves  12  ;  but  18  —  4-|-  2  signifies  that  4  is 
to  be  taken  from  1 8,  and  2  added  to  the  remainder,  which  makes  16. 


18  —  4 

7-16    - 


-7       how  many? 
3  _|_  8  ==how  many' 


7- 

12  —  3  —  2 

7-1-8-1-4 
6- 


3-!- 


—  6  -  -  2 


-  6  —  (9  ~  4  -]-  7)  =  how  many? 

4  -|-  7  -|-  6  ==  how  many? 

( 1 0  -[-  7  -|-  6  —  4)  =  how  many? 


i-  ^    |- 7  -|-  6  —  (9  -[-  8)  =  how  many? 

Obs.  6.  Subtraction  is  the  reverse  of  Addition.  Addition  is  find- 
ing the  sum,  and  Subtraction  is  finding  the  difference  of  numbers. 
Now  if  3  taken  from  9  leaves  6,  it  is  evident  that  3  added  to  6  will 
equal  9.     Hence  : 

To  prove  Subtraction — 

Obs.  7.  Add  the  remainder  and  subtrahend  together ^  and  if  their 
sum  is  equal  to  the  minuend  the  work  is  correct. 

Ex.  1.  A  man  having  98  dollars,  paid  52  dollars  for  a  horse. — 
How  much  had  he  left?  Ans.  46  dollars. 

Solution. — 98  is  composed  of  9  tens  and  8  units;  52  is  composed 
of  5  tens  and  and  2  units  :  now  we  cannot  take  units  from  tens,  nor 
tens  from  units,  because  it  takes  10  of  the  one  (units),  to  make  1  of 
the  other  (tens) ;  but  we  can  take  units  from  units,  and  tens  from 
tens.  Thus :  2  units  from  8  units  leave  6  units,  and  5  tens  from  9 
tens  leave  4  tens,  and  4  tens  and  6  units  =  46. 

Or,  we  may  set  the  numbers  down,  the  less  under  the  greater, 


What  does  a  line  drawn  over  two  or  more  numbers  signify?  What  other 
mark  is  used  for-the  same  purpose?  Of  what  is  Subtraction  Ihe  reverse? — 
Show  the  comparison.  How  do  wo  prove  Subtraction?  Why  cannot  we  take 
units  from  tens,  or  tens  from  units?  How  do  we  set  the  numbers  down  to 
subtract?  ''^  hy  do  we  place  units  under  units,  &c.?  (Sect.  II,  Art.  2,  Obs. 
4.)  Where  do  we  commence  to  subtract?  Where  do  we  s*it  the  remain- 
der? 


Art.  3.  SIMPLE    SUBTU ACTION.  19 

and  subtract  one  from  the  other,   commencing  at  the  right  hand. 
Thus  : 

Proof, 

98       62       In  subtracting,   we  say..  2  from  8  leaves  6,    and 

52       46  place  the  6  under  the  2 ;  then  5  from  9  leaves  4,  and 

—       —  place  the  4  under  the  5. 

46       98       Hence — To  subtract  numbers  : 
Obs.  8.     Place  tlie  less  nimiber  under  the  greater,  setting  units  un- 
der units ^  tens  binder  ten^s,   c£*c.,  and  draiv  a  line  beneath.     Commence 
at  the  right  hand.     Take    successively,  each  figure  in  the  lower  line 
from  the  figure  above  it,  and  write  the  difference  below. 

2.  A  man  had  288  sheep,  and  sold  1 84  of  them.  How  many 
had  he  left?  Ans.   104. 

3.  A  man  has  lived  97  years,  and  was  married  when  he  was  25 
years  of  age.     How  many  years  was  he  married?  Ans.  72. 

4.  A  gentleman  having  948  dollars,  paid  away  736  dollars.  How 
many  dollars  had  he  remaining?  Ans.  212. 

5.  One  book  contains  489  pages,  and  another  contains  372  pages. 
How  ihany  more  pages  in  the  one  than  in  the  other?     Ans.   117. 

6.  A  man's  property  is  worth  6894  dollars,  and  his  debts  amount 
to  4080  dollars..     How  much  will  remain  after  paying  his  debts? 

Ans.  2814  dollars. 

Article  3.     Borrow^ing  and  Carrying  in  Subtraction. 

Ex.  1.  A  man  gave  95  dollars  for  a  horse,  and  68  dollars  for  a 
wagon.  How  much  more  did  he  give  for  the  horse  than  for  the 
wagon?  Ans.  27  dollars. 

Solution. — Here  a  difficulty  occurs  ;  for  we  cannot  take  8  units 
from  5  units.  But  95  is  composed  of  9  tens  and  5  units  ;  now  if 
we  take  1  ten  from  the  9  tens,  and  add  it  to  the  5  units,  we  shall 
have  8  tens  and  15  units  ;  then  8  units  from  16  units  leaves  7  units, 
and  6  tens  from  8  tens  leaves  2  tens ;  and  2  tens  and  7  units  =  27. 

2.  Required — the  difference  between  124  and  86. 

In  this  example  we  cannot  take  6  Operation. 

units  from  4  units,  therefore  we  bor-  tens.         units, 

row  1  ten  from  the  2  tens,  and  add        \2Az=i\\  -|-  14 
the  4  units,  making  14  units,  and  14  86  -=    8-|-    6 

—  6  ^=:  8  units.     Again,  we  cannot  

take  8  tens  from   1  ten,  so  we  bor-         Ans.     3  -]-  8  =  38  rem. 
row  the  1  hundred  and  add  it  to  the  1  ten,  which  makes  11  tens; 
then  11  —  8  :=::=  3  tens,  and  3  tens  and  8  units  =38. 

Explain  the  method  of  separating  a  number  into  its  numerical  parts,  as  in 
Examples  1st  and  2ud.  Explain  the  principle  of  borrowing  ten.  When  the 
lower  figure  is  the  largest,  can  we  subtract  without  resolving  the  numberu 
into  their  numerical  parts? 


"20  COl^fMON    ARITHMETIC,  ScCt.   III. 

Obs.  1.  The  learner  will  perceive  in  both  these  examples,  that 
we  add  ten  to  the  figure  of  the  minuend  when  it  is  less  than  the 
corresponding  figure  of  the  subtrahend*  Indeed,  it  cannot  be  oth- 
erwise, when  we  borrow  1  of  the  next  higher  order,  as  1  unit  of  any 
order  is  equal  ^o  10  units  of  the  next  lower  order.  (Sect.  1.  Art. 
2,  Obs.  3.)     This  is  called  borrowing  ten. 

3.  Required — the  difference  between  626  and  328, 
Operation.         In  the  solution  of  this  example,  instead  of  resolving 
526         the  numbers  into  their  numerical  parts,  as  above,  we 
328  will  set  them  down  according  to  Obs.  8.,  Art.  2.     To 

take  8  from  6  is  impossible ;  therefore  we  will  borrow 

198  1  {itn)  from  the  2  (tens),  and  add  it  to  the  6,  making 

16;  then  16 — 8^=8.  Now  to  compensate  for  the  one  we  bor- 
rowed, we  will  add  1  to  the  next  figure  of  the  subtrahend,  (which  is 
the  same  as  taking  one  from  the  next  figure  of  the  minue7id,)  and  2 
-|-  1  =r  3;  then  2  —  3  is  impossible;  hence  we  will  borrow  1  (hun- 
dred), and  add  it  to  the  2  (tens),  making  12  (tens),  and  12 —  3=9. 
Then3-|-1  =  4,  and5  — 4=rl.  Ans.   198. 

Remark  1. — We  add  1  to  the  next  figure  of  the  subtrahend,  because  it  is 
more  convenient  than  to  take  1  from  the  minuend. 

2. — The  reason  of  carrying  1  is  evident  from  this  fact:  As  we  borrow  1 
from  the  minuend,  we  must  either  make  the  figure  of  the  minuend,  [^which  we 
borrowed  from)  1  less,  or  the  figure  of  the  subtrahend  i7nmediately  beneath  it 
1  greater,  to  pay  for  that  we  borrowed. 

From  the  foregoing  remarks  and  illustrations,  we  derive  the  fol- 
lowing 

GENERAL  RULE  FOR  SUBTRACTION. 

I.  Write  the  less  number  under  the  greater,  that  ihf  figures  of 
the  same  order  may  stand  zinder  each  other*     (Art.  2,  Obs.  8.) 

II.  Commence  at  the  right  hand.  Take  successively  each  figure 
of  the  lower  number  from  the  one  abcrve  it,  and  write  the  remainder 
below.     (Art.  2,  Obs.  8.) 

III.  When  the  figure  in  the  lower  number  is  the  largest,  add 
ten  to  the  figure  above  it,  after  which  subtract  as  usual,  remember- 
ing to  add,  or  carry  1  to  the  next  figure  in  the  lower  number  be- 
fore the  next  subtraction.     (Art.  3,  Obs.  1.  Rem.) 

Proof. — Add  the  remainder  and  subtrahend  together ;  if  their 
sum  is  equal  to  the  minuend,  the  work  is  correct.     (Art.  2,  Obs.  7.) 

EXERCISES   FOR    THE    SLATE. 

1.  A  man  had  1000  dollars,  and  paid  away  478  dollars.  How 
many  dollars  had  he  left?  Ans.  522.  ^ 

2.  A  man  bought  a  farm  for  7864  dollars,  and  afterwards  sold  it 

Explain  the  process.  Why  do  we  add  1  to  the  next  figure  of  the  subtra- 
hend? Explain  the  reason  of  carrying.  What  is  the  general  rule  for  Sub- 
traction ?     The  proof? 


Art.  8.  SIMPLE   SUBTRACTION.  21 

for  975  dollars  less  than  he  gave  for  it.     For  how  much  did  he  sell 
it?  Ans.  6889  dollars. 

3.  A  merchant  bought  8473  dollars  worth  of  goods,  and  sold 
them  for  10312  dollars.     How  many  dollars  did  he  gain? 

Ans.   1839. 

4.  A  man  paid  6767  dollars  for  land,  and  12843  dollars  for  mer- 
chandise. Ilow  much  more  did  he  pay  for  the  merchandise  than 
for  land?  Ans.  6076  dollars. 

5.  A  merchant  sold  goods  to  the  amount  of  15784  dollars;  he 
paid  2897  dollars  less  for  them.     How  much  did  they  cost  him? 

Ans.   12887  dollars. 

6.  America  was  discovered  in  1492.  How  many  years  since,  it 
now  being  1849?  Ans.  357. 

7.  The  United  States  declared  their  Independence  in  the  year 
1776.     How  many  years  since,  the  present  year  being  1849? 

Ans.  73. 

8.  A  gentleman's  income  is  4742  dollars  a  year,  and  his  expenses 
are  3953  dollars  a  year.     How  much  does  he  save  per  year? 

Ans.  789  dollars. 

9.  A  farmer  raised  1747  bushels  of  grain  one  year,  and  1699 
bushels  the  nex'.  year.  How  many  bushels  did  he  raise  the  first 
year  more  than  the  second?  Ans.  48. 

10.  At  a  certain  school  are  423  students,  of  which  137  are  young 
ladies.     How  many  gentlemen  are  there?  Ans.  286. 

11.  From  Columbus,  (Ohio,)  to  Cincinnati,  by  way  of  Springfield 
it  is  127  miles  ;  and  from  Springfield  to  Cincinnati  it  is  85  miles.-  - 
How  far  is  it  from  Columbus  to  Springfield?  Ans.  42  miles. 

12.  In  the  year  1800,  the  population  of  the  United  States  was 
5305925,  and  in  1840  it  was  17069453.  Required— the  increase 
of  population  in  40  years.  Ans.   11763528. 

13.  In  1830  the  population  was  12861192.  Required — the  in- 
crease in  ten  years.  Ans.  4208261. 

14.  A  drover  having  1468  sheep,  sold  948  ;  he  then  bought  467, 
and  sold  987.     How  many  had  he  left?  Ans.  None. 

15.  A  vessel,  the  cargo  of  which  was  valued  at  84000  dollars, 
in  a  storm  lost  part  of  her  cargo  valued  at  272 1 2  dollars.  What 
was  the  value  of  the  remaining  part?  Ans.  56788  dollars. 

16.  A  man  bought  58  dollars  worth  of  wheat,  97  dollars  worth 
of  pork,  73  dollars  worth  of  cheese,  and  a  fine  horse  and  buggy 
for  347  dollars.  He  gave  his  note  for  497  dollars,  and  paid  the  rest 
in  money.     How  much  money  did  he  pay?  Ans.  78  dollars. 

17.  A  man  borrowed  at  one  time  217  dollars,  at  another  time  313 
dollars,  and  at  another  time  428  dollars ;  he  afterwards  paid  869 
dollars.     How  much  did  he  then  owe?  Ans.  89  dollars. 

18.  A  man  at  his  death  left  each  of  his  two  sons  5732  dollars, 
and  each  of  his  three  daughters  4784  dollars,  and  his  widow  the 


f2!  COMMON    ARlTHME'nC.  Scct.  III. 

balance  of  his  property.     How  much   did  the  widow  receive,  the 
estate  being  worth  40000  dollars?  Ans.   14184  dollars. 

19.  A  gentleman's  property  was  worth  34768  dollars  ;  but  a 
store  worth  4762  dollars,  and  14796  dollars  worth  of  goods  were 
destroyed  by  fire.  How  much  had  he  left?     Ans.   15210  dollars. 

20.  A  man  owning  4821  acres  of  land,  gave  one  son  623  acres 
to  another  son  427  acres,  and  to  another  son  873  acres.  How 
many  acres  liad  he  left?  Ans.  2898. 

21.  A  man's  income  is  6000  dollars  a  year.  He  spends  372  dol- 
lars for  clothing,  724  dollars,  for  house  rent,  892  dollars  for  pro- 
visions, 429  dollars  for  servants,  and  527  dollars  for  traveling. — 
How  much  has  he  left  at  the  end  of  the  year?    Ans.  3056  dollars. 

22*  A  man  owing  1 894  dollars,  paid  at  one  time  723  dollars,  at 
another  time  674  dollars,  and  at  another  time  500  dollars.  How 
did  the  account  then  stand?  Ans.  He  overpaid  3  dollars. 

23.  In  1840  the  population  of  the  New  England  States  was 
2234822  ;  of  the  Middle  States  4604345 ;  of  the  Southern  States, 
including  Florida,  5122320  ;  and  of  the  Western  States,  including 
Wisconsin  and  Iowa,  5058154.  Required — the  excess  of  the  pop- 
ulation of  the  Southern  and  Western  States  over  that  of  the  New 
England  and  Middle  States?  Ans.  3341307  inhabitants. 

24.  In  the  last  question,  required — the  excess  of  the  population 
of  the  Middle  States  over  that  of  the  New  England  States. 

Ans.  2369523. 

25.  Also,  required — the  excess  of  the  population  of  the  South- 
ern States  over  that  of  the  Western  States.  Ans.  64166. 

26.  By  the  last  census,  the  population  of  the  Northern  States* 
was  9807007,  and  the  population  of  the  Southern  States*  was 
7256346.  Required — the  excess  of  the  population  of  the  North- 
ern over  that  of  the  Southern  States.  Ans.  2550661. 

27.  A  man  bought  27  dollars  worth  of  sugar,  19  dollars  worth  of 
spice,  43  dollars  worth  of  coffee,  and  10  dollars  worth  of  tea.  He 
gave  in  payment  a  one  hundred  dollar  bill.  Had  he  ought  to  re- 
ceive any  thing  back?     If  so,  how  much? 

Ans.  He  should  receive  1  dollar. 

28.  A  certain  store-house  contained  5970  bushels  of  wheat,  3752 
bushels  of  corn,  5978  bushels  of  rye,  and  9847  bushels  of  oats. 
It  caught  fire,  and  19768  bushels  of  grain  were  saved.  How  many 
bushels  were  destroyed?  Ans.  5779. 

29.  A  ship  of  war  sailing  with  650  men,  lost  in  one  battle  29 
men,  in  another  37,  and  by  sickness  19  more.  How  many  were 
still  living  ?  Ans.  565. 

30.  Four  men  bought  a  lot  of  land  for  978  dollars.  The  first 
man  paid  386  dollars  ;  the  second  paid  97  dollars  less  than  the 

*In  this  example,  by  the  Northern  are  aneant  the  Free  States,  and  by  the  Southern  are 
meant  the  Slave  States. 


SIMPLE   MULTIPLICATION.  23 

first  ;  the  third  paid  73  dollars  less  than  the  second  ;  and  the  fourth 
paid  the  balance.  How  much  did  the  second,  third  and  fourth  men 
pay?  i  The  second  man  paid  289  dollars. 

Ans.      <     "    third  **         216 

(     *'    fourth  "  87 

31.  From  six  trillions,  ei^^hty- seven  millions,  twelve  thousand  and 
three,  take  twenty  billions,  one  hundred  millions,  two  hundred  and 
sixteen  thousand  and  nine.  Ans.  5979986795994. 


SECTION  IV. 
SIMPLE  MULTIPLICATION. 

Article    1.      Mental    Exercises. 

1.  If  one  lemon  cost  4  cents,  how  much  will  3  lemons  cost? 
Solution. — If  one  lemon  costs  4  cents,  2  lemons  will  cost  4-|-4 

=  8  cents,  and  three  lemons  will  cost  8-|-4,  or  4-|-4-|-4=  12 
cents.  Ans.   12  cents. 

2.  What  costs  4  oranges  at  6  cents  apiece? 
6  -|-  6  -|-6  -|-  6  =1^  how  many? 

3.  At  2  dollars  a  pair,  what  will  3  pair  of  boots  cost? 

4.  In  one  yard  are  3  feet;  how  many  feet  in  4  yards? 

5.  How  much  will  5  yards  of  ribon  cost  at  8  cents  a  yard? 

6.  How  much  will  7  lead  pencils  cost  at  5  cents  apiece? 

7.  What  cost  9  pounds  of  saleratus  at  6  cents  a  pound? 

8.  At  10  cents  a  pound,  how  much  will  10  pounds  of  sugar  cost? 

9.  What  cost  8  slates  at  6  cents  apiece? 

10.  At  4  dollars  a  barrel,  how  much  will  7  barrels  of  flour  cost? 

1 1 .  If  one  sheep  costs  2  dollars,  how  much  will  1 1  sheep  cost? 

12.  At  12  cents  a  pound,  how  much  will  6  pounds  of  butter 
cost? 

Article  2.     Definitions,  &c. 

There  is  an  orchard  containing  4  rows  of  trees,  and  6  trees  in  a 
row.     How  many  trees  are  there  in  the  orchard? 

Instead  of  adding  the  4  six  times,  or  the  6  four  times,  as  6 
in  the  preceding  examples,  we  will  shorten  the  operation  by  4 
multiplying  one  by  the  other.     Thus :  — 

24 

To  illustrate  this,  we  will  suppose  these  ****** 
stars  to  be  the  orchard.  We  perceive  ****** 
that  there  are  4  rows  of  stars,  and  6  *****  * 
stars  in  a  row,  and  24  stars  in  all.  There-  ****** 
fore  4  times  6  are  24.  Now  count  the  stars  the  other  way.  We 
perceive  that  there  are  6  rows,  and  4  stars  in  a  row,  and  24  stars 


24 


COMMON    AWTHMETia 


Sect.  IV. 


in  all,  as  before.  Hence,  6  times  4  are  24.  Thus,  we  perceive  it 
makes  no  difference  whether  we  multiply  6  by  4,  or  4  by  6,  the  re- 
sult will  still  be  the  same — ^that  is,  24.  Ans.  24. 

At  5  dollars  a  yard,  what  will  5  yards  of  cloth  cost? 
5  times  5  are  how  many? 

Obs.  1.  This  method  of  repeating  a  number  is  called  Multi- 
plication.    Multiplication  may  therefore  be  defined — 

A  melhod  of  repeating  any  number  a  given  number  of  times. 

Multiplication  may  be  either  of  Simple  or  Compmind  numbers. 

Obs.  2.  Simple  Multiplication  is  tvhen  the  number  to  he  multi- 
plied expresses  things  of  but  one  name  or  kind — as  dollars,  yards, 
&c. 

•  Obs.  3.  The  number  which  we  multiply,  or  which  is  to  be  re- 
peated, is  called  the  Multiplicand. 

The  number  which  we  multiply  by,  or  which  shows  how  many 
times  the  other  is  to  be  repeated,  is  called  the  Multiplier. 

The  result,  or  answer,  is  called  the  Product. 

The  multiplicand  and  multiplier  taken  together  are  called  the 
Factors,  or  producers  of  the  product. 

Obs.  4.  Sign. — The  sign  of  multiplication  is  a  cross  (x)  l^ke 
the  letter  X,  and  signifies  that  the  numbers  between  which  it  stands 
are  to  be  multiplied  together.  Thus:  4X6  signifies  that  4  and 
6  are  to  be    multiplied  together,    the  product   of  which  is  24. 

Before  proceeding  farther,  the  learner  must  commit  accurately 
to  memory  the  following 

MULTIPLICATION  TABLE. 


2  times 

3  times 

4  times 

5  times 

6  times 

7  times 

1  are  2 

1 

are  3 

1 

are  4 

1  are  5 

1  are  6 

lare  7 

2    "    4 

2 

"   6 

2 

"    8 

2  "  10 

2   ' 

*  12 

2   "  14 

3    *'    6 

3 

"    9 

3 

"12 

3  "  15 

3 

"18 

3  "  21 

4    *'    8 

4 

"12 

4 

"16 

4  "  20 

4 

"24 

4   "28 

5    "  10 

5 

"15 

6 

"  20 

5  "  25 

5 

'30 

5  "35 

6    "  12 

6 

*M8 

6 

"  24 

6  "30 

6 

'*36 

6  "  42 

7    **  14 

7 

"21 

7 

"  28 

7  "35 

7 

u  42 

7  "49 

8    "  16 

8 

"24 

8 

"  32 

8  "  40 

8 

'  48 

8  "  5Q 

9    "18 

9 

*'27 

9 

"  36 

9  "  45 

9 

'  54 

9  **  63 

10    **20 

10 

"30 

10 

*«  40 

10  "  50 

10    ^ 

'  60 

10  "  70 

11    "22 

11 

"S3 

11 

"  44 

11    "  55 

11    * 

'•6 

11   «  77 

12    M  24 

12 

"36 

12 

"  48 

12  "  60 

12    ' 

*72 

12  "  84 

What  is  Multiplication?  How  is  it  divided?  What  is  Simple  Multiplica- 
tion? What  is  the  number  we  multiply  called?  What  is  the  number  we  mul- 
tiply by  called?  What  is  the  result,  or  answer,  called?  What  are  the  multi- 
plicand and  multiplier,  taken  together,  called?  What  is  the  sign  o(  multi- 
plication?   What  does  it  signify? 


Art.  3. 


SIMPLE   MULTIPLICATION. 


25 


MULTIPLICATION  TABLE— continubd. 


8  times 

9 

times 

10  times 

1 1  times 

12  times   ] 

1    are 

8 

1 

are 

9 

1 

are    10 

1 

are 

11 

1 

are    12 

2     " 

16 

2 

18 

2 

"     20 

2 

a      24 

3     *' 

24 

•'^ 

27 

3 

"     30 

3 

''     36 

4     « 

32 

4 

36 

4 

"     40 

4 

44 

4 

♦«     48 

5     '•' 

40 

5 

45 

5 

*♦     5 

5 

55 

5 

«     60 

6     " 

48 

6 

64 

6 

"     60 

6 

66 

6 

ic     72 

7     " 

56 

7 

63 

7 

*'     70 

7 

Ti 

7 

^'     84 

8     '' 

64 

] 

7-- 

8 

"     80 

8 

81 

8 

«     96 

9     " 

72 

■) 

■ 

1 

♦*     90 

9 

9^ 

9 

''  108 

10     " 

80 

10 

90 

10 

u    100 

10 

^ 

10 

"  120 

11     - 

88 

11 

99 

11 

"  no 

11 

121 

11 

"   132 

12     ** 

96 

12 

108 

12 

«   120 

12 

132 

12 

«   144 

7X6=  how  many? 
9X5  =  how  many? 
7X4=  '>w  nvniv? 
8X7=  how  many? 
11X6=  how  many? 
12X8=:  how  many? 


Note. — The  pupil  shonlH  not  l«avo  this  Table  until  he  has  thoroughly  com- 
mitted it  to  memory.  Without  knowing  it  perfectly,  no  one  can  becojne  a  good 
arithmetician. 

6X9=  how  many? 
7  X  8  =  how  many? 

9  X   f'  =        »VVlij;inv? 

9X7  =  how  many? 
11  X  12  =  how  many? 
11X7  =  how  many? 

In  each  of  these  questions  we  took  one  number  as  many  times  as 
there  were  units  in  another.     Hence — 

Obs.  5.  To  multiply  one  number  by  another  is  to  take  the  mul- 
ii2)licand  as  many  times  as  there  are  units  in  the  multiplier. 

The  product  of  any  two  numbers  is  the  same,  whichever  factor  is 
taken  as  the  raultipUer. 

Obs.  6.  As  it  does  not  alter  the  name  of  a  number  to  repeat  it, 
and  the  multipUcand  is  always  the  number  repeated,  it  is  evi  lent 
that  the  product  will  always  be  of  the  same  name  as  the  multiplicand.' 

Remark  1. — The  multiplier  must  always  be  an  abstract  number,  as  it  ex 
presses  the  number  of  times  the  multiplicand  is  taken. 

2. — An  abstract  number  is  one  that  has  no  relation  to  any  particular  object 
whatever,  but  merely  expresses  the  number:  as  14,  29,  &,c. 

3 — When  a  number  has  some  relation  to  a  particular  object,  it  is  called  a 
concrete  number  :  as  5  dollars,  8  miles,  &,c. 


To  multiply  one  number  by  another,  is  to  take  the  multiplicand  how  many 
times?  Does  it  make  any  difference  in  the  result  which  factor  is  used  as  the 
multiplier?  Does  it  alter  ^the  name  of  a  number  to  repeat  it?  Which  fac- 
tor is  repeated?  Of  what  name  then  is  the  product?  What  must  the  mul- 
tiplier always  be?  Why?  What  is  an  abstract  number?  A  concrete  num- 
ber? What  numbers  can  be  multiplied  together?  If  the  multiplicand  is 
an  abstract  number,  what  is  the  product?  If  the  multiplicand  ia  a  eoncr«t« 
numbef.  what  is  the  product? 


^6  COMMON     ARITHMETIC.  Sect.  IV. 

4. — As  we  cannot  multiply  by  a  concrete  number,  it  follows  that  we  can 
only  multiply  two  abstract  numbers  together,  in  which  case  the  product  is 
abstract,  or  an  abstract  and  concrete  number  together,  in  which  case  the  pro- 
duct is  concrete. 

Ex.  1.  If  1  yard  of  cloth  cost  5  dollars,  what  will  6  yards  cost? 

Here  we  have  two  concrete  numbers,  viz  :  5  dollars  and  6  yards. 
Kow  the  multiplier  should  be  an  abstract  number,  for  it  would  be 
equally  as  absurd  to  multiply  5  dollars  by  6  yards,  as  it  would  be 
to  multiply  5  houses  by  6  gardens.  But  we  can  resolve  the  question 
in  this  manner:  if  1  yard  costs  5  dollars,  6  yards  will  cost  6 
times  as  muck  as  1  yard,  or  6  iim3s  5  dollars,  loklch  is  30  dol- 
lars.    In  this  way  we  make  the  6  an  abstract  number. 

Let  the  learner  resolve  the  following  questions  in  the  same  man- 
ner : 

2.  A  man  bought  12  cows,  paying  11  dollars  apiece  for  them. 
How  much  did  they  cost  him? 

3    How  much  would  9  calves  come  to,  at  4  dollnrs  apiece?       '^ 

4.  How  much  would  8  pair  of  boots  cost,  at  5  dollars  a  pair? 

5.  How  much  would  32  yards  of  cloth  cost,  at  3  dollars  a  yard? 
Solution. — To  obtain  the  result  in  this  question,  we  set  one  num- 
ber under  the  other,  and  multiply  each  figure  32 

of  the  multiplicand  separately.     Thus —  3 

3  times  2  units  are  G  units,  and  3  times  3  tens  — 

are  9  tens,  and  9  tens  and  6  units  are  96.  Ans.  96  dollars. 

Obs.  7.  Proof. — Take  1  from  the  multiplier ;  multq^ly  the  mul- 
tiplicand hy  the  remainder,  and  to  the  product  add  the  multipli- 
cand ;  if  this  result  is  the  same  as  the  first,  the  ivorh  is  correct. 

Proof  of  the  last  Example. 

32 

3—1=   2 

As  this  result  is  the  same  as  — 

the  first,  the  work  is  supposed  64  .'^ 

to  be  correct,  32 

96 
The  reason  of  this  rule  is  too  evident  to  need  demonstration.-—' 
Let  the  learner  see  if  he  cannot  demonstrate  it  himself. 

6.  Multiply  321  by  4.  Ans.   1284. 

7.  If  1  pound  of  butter  costs  cents,  what  will  16  pounds  cost? 
6  X  8  =  48  =  4  tens  and  8  units ;  we  set  down        Operation. 

the  8  units  ;  then  8  times  1  ten  =  8  tens,  and  4  tens  16 

make  12  tens.  8 

Ans.  128  cents'. 


Art.  2.                                 SLMPLE  MULTIPLICATION.  q27 

This  principle  may  be  illustrated  as  follows  : 
8.  Required — the  product  of  247  by  9. 

1st    Operation.  2d    Operation. 
247                                                   247  =  200  4-40  +  7. 

9  200X9  =  1800. 

40  X  9  =    3G0. 

63  =  product  of  9  into  7  units.  7X9=      63. 

36    =     "         "         '*    4  tens.  


18      =     **         "         "    2  hundreds.  Ans.    2223 


Ans.  2223  =     "         "         "     247.  i^um  woii     " 

By  each  of  these  methods  it  will  be  perceived  that  the  tcn^s  fig- 
ure of  each  partial  product  is  added  to  the  next  partial  product. — 
Thus,  in  63,  the  6  is  added  to  the  next  partial  product,  (36);  and 
in  36,  the  3  is  added  to  the  next  partial  product,  (18.)  Now  as  this 
is  always  the  case,  we  may  shorten  the  operation  by  not  setting 
down  all  the  work,  but  merely  carrying  the  tens,  as  follows  : 

247         Here  we  say  9  times  7  are  63,  and  set  down  the  3, 
9     and  carry  the  6 ;  then,  9  tiines  4  are  36,  and  6  to  car- 

ry  =  42 ;  we  set  down  2  and  carry  4  ;  finally,  9  times 

Ans.  2223     2  are  18,  and  4  to  carry  =  22. 

The  learner  will  perceive  that  each  figure  of  the  product  occu- 
pies the  same  place,  (numerically,)  as  the  figure  multiplied ;  that  is, 
if  the  figure  multiplied  is  units,  the  product  is  units  ;  if  tens,  the 
product  is  tens,  &c.     Hence^ 

Obs.  8.  When  the  multiplier  occupies  the  unit's  pAace,  each  fig- 
ure of  the  product  occupies  the  same  place,  {numerically, )  as  the 
figure  multiplied. 

From  the  preceding  illustrations,  we  deduce  the  following 
RULE — When  the  Multiplier  is  less  than  12. 

I.  Write  the  multiplier  under  the  unit  figure  of  the  midtiplicand. 

II.  Commence  at  the  right  hand.  Midtiply  successively  each  fig- 
ure of  the  multiplicand  hy  the  multiplier,  setting  doivn  the  unit  fig- 
ure of  each  product,  and  carrying  the  tens  to  the  next  product,  as 
in  Addition. 

Proof, — Take  1  from  the  multiplier ;  multiply  the  multiplicand 
hy  the  remainder,  and  to^  the  product  add  the  multiplicand ;  if  this 
result  is  the  same  as  the  first,  the  ivork  is  correct.    (Obs.  7.)  , 

Give  the  solution  of  Example  1,  and  show  how  the  multiplier  is  made  au 
abstract  number.  How  do  we  prove  maitiplication?  Explain  the  principle 
of  carrying,  and  show  why  it  is  correct.  Is  there  any  shorter  way  than 
to  set  down  the  whole  product  of  each  figure  multiplied,  or  of  separating 
the  multiplicand  into  its  numericiil  parts?  Explain  it.  What  place  does 
each  figure  of  the  product  occupy  ,  numericaJiy?  What  inference  is  de- 
duced from  this?  What  is  the  Rule  when  the  multiplier  is  less  thau  12? 
The  proof?  ,  ,  /. 


28  COMMON    ARITHMETIC.  ScCt.  IV . 

EXERCISES    FOR    THE    SLATE. 

1.  How  much  would  356  pair  of  shoes  cost,   at  2  dollars  a  pah*1 

Ans.  712  dollars. 

2.  How  much  would  431  yards  of  cloth  cost  at  5  dollars  a  yard? 

Ans.  2165  dollars. 

3.  If  a  man  rides  120  miles  in  one  day,  how  tar  can  he  ride  in 
6  days?  Ans.  720  miles. 

4.  In  one  mile  are  320  rods.     How  many  rods  in  8  miles? 

Ans.  2560. 

5.  How  much  would  9  horses  cost  at  140  dollars  apiece? 

Ans.  1260  dollars. 

6.  How  much  would  398  pounds  of  salaratus  cost,  at  6  cents  a 
pound?  Ans.  2388  cents. 

7.  How  much  would  554  hats  cost,  at  5  dollars  apiece? 

Ans.  2770  dollars. 

8.  There  are  8  quarts  in  a  peck.  How  many  quarts  in  389 
pfecks?  Ans.  3112. 

9.  How  much  would  674  yards  of  cloth  cost,  at  4  dollars  a  yard? 

Ans.  2696  dollars. 

10.  How  much  would  6  carriages  cost,  at  494  dollars  apiece? 

Ans.  2964  dollars. 

11.  How  much  would  738  barrels  of  flour  cost,  at  6  dollars  a  bar- 
rel? Ans.  4428  dollars. 

12.  At  8  shillings  a  day,  how  many  shillings  can  a  man  earn  in 
325  days?  Ans.  2600. 

13.  How  many  yards  will  12  pieces  of  cloth  contain,  each  piece 
containing  42  yards?  Ans.  504. 

14.  How  much  would  the  above  cloth  cost,  at  9  cents  a  yard? 

Ans.  4536  cents. 

15.  How  much  would  386  hats  cost,  at  8  dollars  apiece? 

Ans.  3088  dollars. 

16.  How  much  would  273  pair  of  boots  cost,  at  5  dollars  a  pair? 

Ans.   1365  dollars. 

17.  How  much  would  12  stoves  cost,  at  37  dollars  apiece? 

Ans.  444  dollars. 

18.  How  much  would  198  pounds  of  coffee  cost,  at  11  cents  a 
pound?  Ans.  2178  cents. 

19  How  much  would  177  pounds  of  sugar  cost,  at  7  cents  a 
pound?  Ans.   1239  cents. 

20.  How  much  would  789  acres  of  land  cost,  at  12  dollars  an 
acre?  Ans.  9468  dollars. 

Article  3.     When  the  Multiplier  exceeds  12. 
fgl    Ex.  1.  What  will  243  acres  of  land  cost,  at  24  dollars  an  acre? 
Solution. — ^We  meet  with  a  difficulty  in  this  example,  as  our  mul- 


Art.    3.  SIMPLE   MULTIPLICATION.  28i  * 


rrt    0 


tiplier  is  greater  than  12.  But  24  is  composed  of  2  tens  and  4 
units. 

We  will  therefore  work  this  question  as  follows  : 

In  the  first  place  we  multiply  by  the      Ojyeratlon. 
4  units,  as  usual.     We  next  multi-  243 

ply  by  the  2  tens.     Now  in  reality,  24 

we  multiply  by  20,  because  2  tens  

=  20.     Then  20  times  3  are  60  ;  972  =  product  by  4  units, 

or,  if  we  reject  the  cipher,  2  times         486    =         **  2  tens. 

3  are  6  ;  but  as  the  2  is  2  tens,  the         

6  is  also   6  tens,  and  should  there-         5832  =  "  24 

fore  occupy  the  ten's  place,  the  same  as  if  the  cipher  were  added. 
We  multiply  the  remaining  figures  as  usual.  Finally,  we  add  the 
two  products  together,  (because  20  -]-  4  =  24,)  as  they  stand,  and 
their  sum  is  the  product  required. 

Remark.— In  writing  our  factors,  we  place  figures  of  the  same  order  under 
each  other. 

2.  Multiply  5297  by  642. 

The  6  in  the  multiplier,  in  this  example,  occu-  Operation. 

pies  the  hundred's  place,  and  to  multiply  by  it  is  5297 

the  same  as  to  multiply  by  600.     Therefore,  we  642 

place  the  first  figure  of  the  product  in  the  hun-  

dred's  place.      Otherwise,  we  proceed  as  in  the  10594 

last  example.  21188 

31782 


Ans.  3400674 

Obs.  1.  From  the  preceding  illustrations  we  perceive  that  the  first 
figure  of  each  p)artial  product  occupies  the  same  place,  numerically, 
as  tlie  figure  by  ichich  we  multiply.  Hence — When  the  multiplier 
exceeds  12  : 

Obs.  2.  We  first  multiply  by  each  figure  ofi  the  multiplier  sepa- 
rately, remembering  to  place  the  first  figure  of  the  product  directly 
under  the  figure  by  which  we  multiply ;  and  then  add  together  the 
several  partial  products,  as  they  standi  for  the  required  product. 

3.  Multiply  49832  by  148.  Ans.  7375136. 

4.  How  much  would  1927  tons  of  iron  cost,  at  108  dollars  a  ton? 


Explain  the  solution  of  Example  1st,  and  show  the  several  steps  of  the 
operation.  What  place,  numerically,  does  the  first  figure  of  each  partial 
product  occupy?  Give  the  rea.«on  of  this.  How  do  we  proceed  when  the 
multiplier  exceeds  12?  When  there  are  ciphers  between  the  significant  fig- 
ures of  the  multiplier  how  do  we  proceed?  What  is  the  general  rule  for 
Multiplication?     The  proof? 


30  COMMON    ARITIiMLTIC,  ScCt.    IV. 

Operation, 
In  this  example  there  are  no  tens,  but  as  1927 

0  multiplied  into  any  number  produces  only  108 

0,  we  may  omit  the  0,  and  multiply  by  the  

significant  figures,  only,  observing  to  place  15416 

the  first  figure  of  each  partial  product  accor-  1927 

ding  to  Obs.  2.     Hence —  

Ans.  208116  dollars. 

When  there  are  ciphers  between  the  significant  figures  of  the 
multiplier: 

Obs.  3.  We  multiphj  by  the  sigwficant  figures  only,  and  in  all 
oth^r  respects  proceed  according  to  Obs.  2. 

5.  Mukiply  1776  by  305.  Ans.  541680. 

From  the  preceding  remarks  and  illustrations,  we  derive  the  fol- 
lowing 

GENERAL  RULE  FOR  MULTIPLICATION. 

I.  Write  the  multiplier  under  the  multiplicand,  placing  figures 
of  the  same  order  under  each  other. 

IL  Commence  ivith  the  units,  and  multiply  each  figure  of  the 
multiplicayid  by  each  significant  figure  of  the  multiplier,  placing  the 
first  figure  of  each  partial  product  directly  under  the  figure  by 
which  we  multiply.     (Obs.  2.) 

III.  Add  together  the  several  partial  products  as  they  stand ; 
their  sum  will  be  the  product  required.     (Obs.  2.) 

Proof. — Take  1  from  the  multiplier ;  multiply  the  multiplicand 
by  the  remainder,  and  to  the  product  add  the  multipMcand ;  if  this 
residt  is  the  same  as  the  first,  the  work  is  correct.     (Art.  2,  Obs.  7.) 

EXERCISES    for    THE    SLATE. 

1.  How  much  would  596  sheep  cost,  at  3  dollars  apiece? 

Ans.  1788  dollars. 

2.  How  much  would  1487  yards  of  cloth  cost,  at  5  dollars  a 
yard?  Ans.  7435  dollars. 

3.  How  much  would  276  horses  cost,  98  dollars  apiece? 

Ans.  27048  dollars. 

4.  How  much  would  138  acres  of  land  cost,  as  47  dollars  an 
acre?  Ans.  6486  dollars. 

5.  How  much  would  976  yoke  of  oxen  cost,  at  67  dollars  a  yoke? 

Ans.  65392  dollars. 

6.  How  much  would  235  tons  of  hay  cost,  at  16  dollars  a  ton? 

Ans.  3760  dollars. 

7.  How  much  would  2798  acres  of  land  cost,  at  23  dollars  an 
acre?  Ans.  64354  dollars. 

8.  If  a  man  can  raise  47  bushels  of  corn  on  one  acre  of  land,  how 
many  bushels  can  he  raise  on  179  acres?  Ans.  8413. 


Art.    3.        -'  SIMPLE    MULTIPLICATION.  31 

9.  How  much  would  329  carnages  come  to,  at  574  dollars  apiece? 

Ans.  188846. 

10.  How  many  days  has  a  man  lived  who  is  67  years  old,  allow- 
ing 365  days  to  the  year?  Ans.  24455. 

1 1 .  Allowing  he  has  lived  but  59  years,  how  many  days  would  it 
be?  Ans.  21635. 

1 2.  If  he  has  lived  84  years,  how  many  days  is  it? 

Ans.  30660. 

13.  If  it  takes  872  men  204  days  to  dig  a  canal,  how  long  will 
it  take  1  man  to  dig  hi  Aiis.    177888  days. 

14.  How  much  would  279  horses  cost,  at  207  dollars  apiece? 

Ans.  57753  dollars. 

15.  How  much  would  tlM3  above  horses  cost,  at  196  dollars 
apiece?  Ans.  54684  dollars. 

16.  How  much  would  they  cost,  at  175  dollars  apiece? 

Ans.  48825  dollars. 

17.  How  many  dollars  would  4072  men  receive,  if  each  received 
408  dollars?  Ans.  1661376. 

18.  If  a  vessel  should  sail  129  miles  a  day,  how  many  miles 
would  she  sail  in  437  days?  Ans.  56373  miles. 

19.  How  much  would  19407  pounds  of  opium  cost,  at  52  shil- 
lings a  pound?  Ans.  1009164  shillings. 

20  How  much  would  96  thousand  feet  of  boards  cost,  at  13  dol- 
lars a  thousand?  Ans.   1248  dollars. 

21.  How  much  would  284  hogsheads  of  molasses  cost,  at  29  dol- 
lars a  hogshead?  Ans.  8236  dollars. 

22.  If  47  men  should  form  a  partnership,  each  man  paying  7684 
dollars,  what  amount  of  capital  would  be  invested? 

Ans.  361148  dollars. 

23.  How  much  would  923  pieces  of  broadcloth  cost,  at  312  dol- 
lars a  piece?  Ans.  287976  dollars. 

24.  How  much  would  867  town  lots  cost,  at  489  dollars  apiecel 

Ans.  423963  dollars.    ^^^ 

25.  Suppose  476  men  were  to  receive  237  dollars  apiece  for  do- 
ing a  pit  ce  of  work,  how  much  would  they  all  receive? 

Ans.   112812  dollars. 

26.  If  a  carriage  wheel  turn  over  348  times  in  going  a  mile,  how 
many  times  will  it  turn  over  in  going  957  miles? 

Ans.  333036  times. 

27.  Multiplyone  hundred  and  two  thousand,  four  hundred  and 
seven,  by  threethousand  and  seventeen.  Ans.  308961919. 

28.  Multiply  four  hundred  and  ninety-six  thousand,  five  hundred 
and  fourteen,  by  thirty-three  thousand,  nine  hundred  and  ninety- 
nine.  '  Ans.  16880979486. 

29.  Multiply  two  million,  seven  hundred  and  five,  by  six  hun- 
dred and  seventy-two  thousand,  two  hundred  and  fifteen. 

Ans.  1344903911675. 


32  COMMON    ARITHMETIC.  Sect.    IV.    . 

30.  Multiply  sixty-six  trillion,  six  hundred  and  sixty-six  billion, 
six  hundred  and  sixty-six,  by  one  hundred  and  forty-four  million, 
one  hundred  and  forty-four. 

Ans.  9599913599999904095904.       ? 

Article  4.     Contractions  in  Multiplication.  , 

Case  l.—  When  the  Multiplier  is  10,  100,   1000,  &c.— 

Ex.  1.  How  much  must  I  pay  for  47  cows,   at  10  dollars  apiecel 

Ans.  470  dollars. 

As  fimry  removal  of  a  figure  townrda  thp  Ipft  /.ujtK.i^fis-  its  value 
ten  times f  (Sect.  1.  Art.  2.  Obs.  3.)  it  follows,  that  to  multiply  any 
number  by  10,  we  need  only  remove  it  one  place  to  the  left,  or  sim^ 
ply  annex  a  cipher;  to  multiply  by  J 00,  remove  it  two  places, 
or  annex  two  ciphers  ;  to  multiply  by  1000,  remove  it  /^ree  places,  or 
annex  three  ciphers,  &c.  Hence — To  multiply  by  1  with  any  num- 
ber of  ciphers  annexed : 

Obs.  1.  Annex  to  the  multiplicand  as  many  ciphers  as  there  are 
at  the  right  of  the  multiplier* 

Note. — Annex  means  to  place  after,  or  at  the  right  hand. 

2.  There  are  10  cents  in  one  dime  ;  how  many  cents  are  there  in 
327  dimes? 

3.  There  are  100  cents  in  one  dollar;  how  many  cents  in  385  dol- 
lars? 

4  There  are  1000  mills  in  one  dollar;  how  many  mills  in  476 
dollars? 

5.  There  are  10000  mills  in  an  eagle  ;  how  many  mills  in  835 
eagles? 

6.  How  many  pages  are  there  in  575  books,  each  book  having 
1000  pages? 

Case  2. — Multiplication  hy   Comp)osite  Numbers. 

Obs.  2.  A  Composite  number  is  a  number  which  can  be  ob- 
tmned  by  multiplying  two  or  more  factors  together,  each  of  which 
is  greater  than  unity.  Thus,  12  can  be  obiaiued  by  multiplying  3 
and  4,  or  2  and  6  together;  therefore,  12  is  a  composite  number. 
Also,  24  is^composed  of  8  X  3,  or  4  X  6,  or  12  X  2,  or  4  X  2  X  3, 
or  2  X  2  X  2  X  3,  either  of  which  expressions  multiplied  together 
will  produce  24.     Hence,  24  is  a  Composite  number. 

Is  48  a  composite  number?  Is  49?    54?    81?    120?    144?    132? 

Name  the  factors,  or  parts,  of  each  of  the  above  numbers. 

The  learner  will  find  all  these,  as  well  as  all  the  composite  num- 

What  is  it  necessary  to  do  to  multipiy  by  10?  Why?  By  100?  Why? 
By  1000?  Why?  How  then  can  we  multiply  by  1  with  any  number  of 
ciphers  annexed?  What  does  annex  mean?  What  is  a  Composite  number? 
Give  exa,r.plcfe.  Wili  any  number  admit  of  mora  than  one  set  of  factors? 
Name  some  such  numbers. 


Art.  4.  CONTRACTIONS   tN    MCLTil^LlOA JIOK^  S3 

bers  followinor,  in  the  Midtlpllcailon  Table,  if  ho  examines  it  atten- 
tively,, t^ge  the  r  with  oai:  st^tof  factors  of  whicli  they  are  composed. 
Many  of  them,  ho vvc'ver,  will  acifi)it  of  other  factors  than  those 
named  in  the  Tablu.  Let  the  karncr  see  if  lie  cannot  ascertain 
what  they  a^-e. 

Obs.  3.  TJiB  numbers,  or  factors,  which  are  muliiplied  together 
to  produce  any  number,  are  called  the  component  parts  of  that  num- 
ber. Thus  :  3  and  4  are  the  component  piu'ts  of  12,  because  3x4 
=  12  ;  and  3,  2  and  5  are  the  component  parts  of  thirty,  because 
3  X  2  X  5  =  30. 

What  are  the  component  parts,  or  factors,  of  36?  42?  b^l  72? 
84?  90?  96?   100?   121?   132?  144? 

Resiark  1. — A  number  which  cann«t  be  produced  by  the  multiplication  of  two 
or  more  factors,  is  called  a  prime  number.  Thus  :  2, 3,  5,  7,  11,  &,c,,  are  prime 
numbers. 

2. — Ecery  number  must  be  either  a  prime  or  a  cmnposite  number.  Every  even 
number  is  a  composite  nun){)er.     An  oven   numiter  always  ends  with  0,  2,  4, 

6,  or  8,  and  can  always  be  divided  by  2 — that  is,  2  is  one  factor  of  all  such 
numbers  All  numbers  ending  with  5,  or  0,  are  composite  also,  because  5  ig 
one  factor  of  all  such  numbers. 

3, — A  prime  number  is  always  odd.  An  odd  number  always  ends  with  1,  3, 
5,  7,  or  9,  and  ctmuot  be  divided  by  2— that  is,  2  is  not  a  factor  of  such  num- 
bers.    A  composiiQ  number  may  be  either  odd  or  even. 

4. — 2  is  a  pri7m  number,  aillrough  it  is  even.  This  is  the  only  exceptien. 
The  learner  will  unders  and  tiie  nature  of  odd  and  even  numbers  better  whea 
he  has  become  acquainted  with  Division. 

Tell  which  are  prime  and  which  are  composite  numbers  in  the  following  :  4, 

7,  12,  15,  17,  19,  20,  21,23,  25,  27,  29,  51,56,  59,  63,  79,  84,  121,  143, 
15G,  108,  120. 

'i'he  following  is  a  list  of  all  the  prime  numbers,  from  1  to  100  :  2,  3,  5,  7, 
9,  11,  13,  17,  19,  23,  29,  31,  37,  41,  43,  47,  53,  59,  61,  67,  71,  73,  79,  83,  89, 
97. 

1.  How  much  would  18  cows  cost,  at  23  dollars  apiece? 
1st    Operation.  2nd    Operation. 

23  23  18=6X3. 

18  <5  one  iactorof  18. 

184  138  =  cost  of  6  cows. 

23  3  the  other  flictor  of  18, 

Ans.   414  dollars.      Ans.  414  =  cost  of  3  times  6,  or  18  cows. 
,    In  this  case  1 8  is  composed  of  two  factors,  6  and  3.     Then  if  we 

What  are  the  component  parts  of  a  number?  Give  examples.  What  is  a 
prime  number?  Give  examples.  What  must  every  number  be?  What  are 
all  even  numberfi?  With  wiiat  doe.'*  an  even  number  always  end?  What  are 
all  numbers  ending  with  5?  What  is  a  prime  number  always?  With  what 
does  au  odd  number  always  end?  What  may  a  composite  laumber  be?  What 
even  number  is*  a  prime?    Are  there  aiiy  otiier  exceptions? 


m  COMMON    ARITHMETIC.  "*'^  Scct.    IV. 

niTiltiply  by  6,  we  shall  find  liow  miidi  G  cows  would  cost.  Now  18 
cows  wiil  evidently  cost  3  times  as  much  6  cows,  because  6X3  = 
18.  Therefore,  we  multiply  the  price  of  6  cows  by  3,  which  gi\  es 
the  price  of  18  cows. 

Hence — T©  multiply  by  a  composite  number  : 

Obs.  4.  Resolve  the  multiplier  into  its  componevi  'parts,  or  fac- 
tors, then  multiply  the  multiplicand  first  hy  one  of  these  parts,  and 
the  product  thence  arising  hy  the  other,  and  so  on.  The  last  pro- 
duct will  be  the  answer  required. 

Remark  1. — It  matters  not  by  wliich  ffictor  we  multiply  first  ;  if  no  mis- 
take is  made,  i\\e  final  result  will  be  the  same.  The  reason  of  this  is,  because 
the  product  of  all  the  factors  of  any  number,  in  whatever  order  taken,  is  equal  to 
that  number. 

2. — The  learner  will  observe  that  there  is  a  difference  between  resolving  a 
number  into /actors,  and  separating^  it  in\o  parts.  The  former  has  reference  to 
Multiplication,  and  the  latter  to  Addition.  The  product  of  the  factors  of  any 
number  is  the  same  as  the  suyn  of  its  parts  ;  and  it  is  an  established  axiom  in 
mathematics,  thai  iht  whole  is  equal  to  the  sum  of  all  its  parts  ;  or,  the  product  of 
all  its  factors.*  Thus:  2  and  8,  or  4  and  4,  or  2  and  2  and  4,  or  2  and  2  and  2 
and  2,  are  the  factors  of  16;  but  14  and  2,  12  and  4,  10  and  6,  3  and  13,  &c., 
are  its  parts. 

2.  How  much  would  26  tons  of  liay  cost,  at  14  dollars  a  ton? 
[14  =  7  X  2.]  Ans.  364  dollars.  ;;; 

3.  At  75  dollars  apiece,  how  much  would  27  horses  cost?  [27  = 
3X3X3.]  Ans.  2025  dollars. 

4.  How  much  w^ould  147  acres  of  land  cost,  at  36  dollars  an 
acre?     36  =  4  X  3  X  3.]  Ans.  5292  dollars. 

5.  How  much  would  41  books  cost,  at  25  shillings  apiece?  [25 
=  5X5.]  Ans.   1025  shillings. 

6.  How  much  would  54  town  lots  cost,  at  378  dollars  apiece?  [54 
=  9X6.]  Ans.  20412  dollars. 

7.  How  much  would  478  sheep  cost,  at  24  shillings  apiece?  [24 
=  6X4.]  Ans.   11472  shillings. 

8.  How  much  would  123  wagons  cost,  at  72  dollars  apiece?  [72 
=  9X8.]  Ans.  8856  dollars. 

Case  3. —  When  there  are  ciphers  at  the  right  of  either,  or  loth 
factors. 

Ex.  1.  How  much  would  64  acres  of  land  come  to,  at  20  dol- 
lars per  acre? 

Operation.  20  is  a  composite  number,  the  factors  of  which 

64  are  2  and  10.     Then  64  X  2  =  128  ;  and  annex- 

20  ing  a  cipher — which  is  the  same  as  multiplying  by 

ten,  (Obs.  1.) — we  have  1280  dollars  as  our  an- 

Ans.  1280  swer. 

How  do  we  multiply  by  a  composite  nu;nber?  Does  it  make  any  differ- 
ence wi»h  the  final  result  which  factor  we  multiply  by  first?  Why  not? 
Is  there  any  difference  between  resolving  a  number  into  factors,  and  sepa- 
rating it  into  parts?    Explain  the  difference,  and  give  examples* 

*  That  Is,  all  of  one  set  of  parts,  or  factoin. 


Art    4.  CONTRACTIONS    IN    MULTIPLICATION.  85 

2.  How  miicli  would  40  horses  cost,  at  60  dollars  apiecel 

Operation.         60  and  40  are  both    composite  numbers.     The 

40         component  parts  of  the  one  are  6  and   10,   and  of 

60         the  other  4  and   10.     Now  6  X  4  X  10  X  10  is  the 

same  as  6  X  10  X  (4  X  10.)  (Obs.  4,  Rem.    1.) 


Ans.  2400  doll.  Then  6  X  4  =  24,  and  10  X  10  =  100,  and 
24  X  100  =  2400.  (Obs.  1.)  The  learner  will  perceive  that  the 
sig-nificant  figures  only  are  multiplied,  and  the  ciphers  at  the  right 
of  both  factors  are  merely  annexed  to  the  product.  He  will 
also  perceive  that  10  is  one  factor  of  all  numbers  ending  with  a  ci- 
pher, and  the  significant  figures  another  factor.  Hence — 
When  there  are  ciphers  at  the  right  of  either  or  both  factors  : 
Obs.  5.  Multiply  the  significant  figures  together,  only,  and  to 
the  product  annex  as  many  ciphers  as  there  are  at  the  right  of  both  fac- 
tors. 

3.  There  are  50  cents  in  a  half  dollar.  How  many  cents  in 
320  half  dollars?  Ans.   16000. 

4.  There  are  320  rods  in  a  mile.  How  many  rods  are  there  in 
1820  miles?  "     Ans.  582400. 

5.  How  much  would  3900  pounds  of  tea  come  to,  at  80  cents  per 
pound?  Ans.  312000  cents. 

6.  An  army  of  13000  men  having  plundered  a  city,  each  man^s 
share  of  the  spoil  was  480  dollars.  What  was  the  amount  of  plun- 
der taken?  Ans.  6240000  dollars. 

7.  There  are  160  square  rods  in  an  acre  ;  how  many  square  rods 
in  3900  acres?  Ans.  624000. 

8.  How  many  pages  would  there  be  in  270  books,  if  there  were 
360  pages  in  each?  Ans.  97200. 

9.  Sound  moves  at  the  rate  of  1130  feet  per  second ;  how  far 
will  it  move  in  4716  seconds?  Ans.  5329080  feet. 

10.  In  a  certain  building  there  are  24  rooms  ;  12  of  these  rooms 
have  one  window  each,  and  12  have  2  windows  each  ;  each  window 
has  24  lights.     How  many  lights  in  the  building?         Ans.  864. 

11.  It  has  been  computed  that  an  elm  produces  annually,  at  an 
average;  three  hundred  and  twenty-nine  thousand  grains,  or  seeds, 
each  of  which  would  produce  a  tree.  How  many  trees  might  be 
produced  from  178  elms,  in  97  years?  Ans.  5680514000.   ^ 

12.  Multiply  956802700  by  324007020. 

Ans.  310010781  £54954030. 

13.  Multiply  1234567890  by  9876543210. 

Ans.  12193263111263526900. 


Of  what  numbers  is  lO  always  a  factor?  What  is  another  factor?  When 
there  are  ciphers  at  the  right  of  eilher,  or  both  factors,  how  do  we  proceed? 
Explain  this. 


M  COMMOK    ARITHMETIC,  Scct.    T. 

14.  Multiply  two  million,  three  hundred  and  fifty-seven  thousand, 
by  (AVu  hunorea  mi^  luu  teen  thousand,  tliree  hundred. 

Ans.  506105100000. 

15.  Multiply  seven  hundred  and  seventy-seven  thousand,  sev- 
hundred  ajid  seventy,  by  fifty-five  thousand,  five  hundred  and  fifty. 

Ans.  43205123500. 

16.  Multiply  eight  hundred  million,  nine  hundred  and  forty 
thousand,  three  hundred,  by  twenty  million,  one  hundred  and 
forty  thousand,  six  hundred. 

Ans.  16131418206180000. 


SECTION  V. 
SIMPLE  DIVISION. 

Article  1.     Mental  Exercises. 

1 .  At  4  cents  apiece,  how  many  lemons  can  you  buy  for  1 2  cents? 
Solution. — The  first  lemon  costs  4  cents,  and  you  have  12  —  4  = 

8  cents  left.  The  second  lemon  costs  4  cents,  and  you  have  8  —  4 
=  4  cents  left,  with  which  you  can  buy  but  1  lemon.  Therefore, 
you  can  buy  3  lemons  for  12  cents. 

2.  How  many  oranges,  at  5  cents  apiece,  can  you  buy  for  15 
cents? 

5  is  one  factor  of  15  ;  what  is  the  other  factor?     (See  Multi- 
plication Table.)  Ans.  3. 

3.  How  many  hats,  at  5  dollars  apiece,  can  be  bought  for  20  dol- 
lars? 

4.  How  many  yards  of  c'oth  can  be  bought  for  24  dollars,  at  4 
dollars  a  yard? 

6.  At  3  shillings  a  bushel,  how  many  bushels  of  potatoes  may 
be  bought  for  1 8  shillings? 

6.  At  6  shillings  a  pair,  how  many  pair  of  gloves  may  be  bought 
for  36  shillings? 

7.  At  3  dollars  apiece,  how  many  sheep  can  be  bought  for  33 
dollars? 

8.  At  7  shillings  apiece,  how  many  books  may  be  bought  for  49 
shillings? 

9.  If  a  pound  of  tea  costs  7  shillings,  how  many  pounds  may  be 
bought  for  63  shillings? 

10.  If  I  pay  6  cents  a  mile  for  riding  on  the  stage,  how  many 
miles  can  I  ride  for  54  cents? 

1 1.  In  owe  peck  are  8  quarts.    How  many  pecks  are  there  in  72 
quarts? 

12.  ThferiB  are  7  days  in  a  weiek  ;  how  many  weeks  are  there  in 
42  dhf^f 


Art.  2. 


«IMPLE   mViSiON. 


Article  2.     Definitions,   .fee. 

The  separ.  ting  a  number  into  equal  parts,  as  in  the  preceding 
examples,  is  called  Division.     Hence — 

Obs.  1.  Division  may  be  defined — the  sepcraiivg  a  ni-mher 
into  any  given  number  of  equal  parts  ;  or,  finding  how  often  one 
number  is  contained  in  another, 

Diviiion  may  be  either  of  Simple  or  Compound  numbers. 

Obs.  2.  Simple  Division  is  when  the  number  to  be  divided  ex- 
presses things  of  but  one  name,  or  kind — as  dollars,  bushels  <fec. 

Obs.  3.     The  numher  which  is  to  be  divided  is  called  the  Dividend. 

The  number  we  divide  by  is  called  the  Divisor. 

The  result,  or  answer  is  called  the  Quotient,  which  is  derived  from 
the  Latin  word  quoties,  signifying  how  many. 

Obs.  4.  Sign. — The  sign  of  Division  is  a  short  horizontal  line 
between  two  dots,  (-f-)-  It  shows  that  the  number  before  it,  is  to 
be  divided  by  the  number  after  it.  Thus:  24 -j- 3  shows  that  24 
is  to  be  divided  by  3. 

a.  Division  is  also  sometimes  expressed  by  writing  the  divisor 
under  the  dividend,  with  a  line  between  them ;  thus  :  \*  shows 
that  24  is  to  be  divided  by  3,  cs  before. 

Before  proceeding  farther,  the  learner  must  commit  accurately  to 
memory  the  following 


DIVISION 

TABLE. 

2 

in 

3  in 

4 

in 

5 

in 

6 

n 

7  in 

2 

once 

3  once 

4 

once 

5  once 

6  once 

7  once 

4 

2 

6 

2 

8 

2 

10 

2 

12 

2 

14   2 

6 

3 

9 

3 

12 

3 

15 

3 

18 

3 

21   3 

8 

4 

12 

4 

16 

4 

20 

4 

24 

4 

28   4 

10 

6 

15 

5 

20 

5 

25 

5 

30 

5 

35   5 

12 

6 

18 

6 

24 

6 

30 

6 

36 

6 

42   6 

14 

7 

21 

7 

28 

7 

35 

7 

42 

7 

49   7 

16 

8 

24 

8 

32 

8 

40 

8 

48 

8 

6Q       8 

18 

9 

27 

9 

36 

9 

45 

9 

54 

9 

63   9 

8 

in 

9 

in 

10 

in 

n  in 

12  in 

8 

once 

9 

0 

nee 

10 

once 

11 

once 

1 2   once 

16 
24 

2 
3 

18 
27 

2 
3 

20 
30 

2 
3 

22" 

2 

24    2 
36    3 

33 

3 

32 

4 

36 

4 

40 

4 

44 

4 

48    4 

40 

5 

45 

5 

50 

5 

bb 

5 

60    5 

48 

6 

54 

6 

60 

6 

%^ 

6 

72    6 

56 

7 

63 

7 

70 

7 

11 

7 

84    7 

64 

8 

72 

8 

80 

8 

88 

8 

96    8 

72 

9 

81 

9 

90 

9 

99 

9 



108    9 

How  »  DiviBloii  defiu^sd  ?     How  divided  ?     What  is  Simple  Diviaton  ? 

-4 


38  COMMON    ARITHMETIC.  Scct.   V. 


2 1  -f-  7  =  how  many  ? 
48  -f-  G  =  how  many  ? 
54  -f-  9  =  how  many  ? 
27  -V-  9  =  how  many  ? 
35-r-  5  =  how  many  ? 
54  -f-  6  =  how  many  ? 


63  -f-  7  =  how  many  ? 
64-j-  8  =  how  many  ? 
32  -f-  4  =  how  many  ? 
49  -f-  7  =  how  many  ? 
24  -r-3  =  how  many  ? 
72  -J-  9  =:  how  many  ? 


Ex.   1.  A  person  paid  369  dollars  for  3  horses  ;  how  much  was 
that  api 


lece 


Solution.— m^  =  300  -|-  60  +  9.  Then  3  in  300  goes  100  times; 
3  in  60  goes  20  times  ;  and  3  in  9  goes  3  times  ;  and  100  +  20  +  3 
=  123.  Ans.   123  dollars. 

We  divide  thus  by  3,  because  it  is  evident  that  if  3  horses  cost 
369  dolldrs,  one  horse  will  cost  one-third  as  much  as  3  horses,  or 
one-third  of  369  dollars. 

But   there  is  a  method  of  dividing  numbers  without  separating 

them  into  their  numerical  parts,  which  we  will  illustrate  as  follows  : 

Operation.  By  this  method  we  place  the  divisor  at 

Divisor  Dividend  the  left  of  the  dividend,  with  a  line  between 

3  )  369  them.     We  next  draw  a  line  below  the  div- 

dend  to  separate  it  from  the  quotient,  -which 

123  Quotient,    is  to  be  placed  below. 

Then  we  say  3  in  3  (hundreds)  goes  1  (hundred)  times,  and  write 
the  1  below  the  3  ;  then  3  in  6  (tens)  goes  2  (tens)  times,  and  we 
write  the  2  under  the  6  ;  then  3  in  9  (units)  goes  3  (units)  times, 
and  we  write  the  3  under  the  9,  and  the  work  is  done. 

Obs.  5.  The  pupil  will  observe  that  the  form  used  is  of  compar- 
atively little  importance,  it  being  used  merely  for  the  sake  of  coyiveni- 
ence.  In  short,  there  is  no  form  by  which  any  operation  in  Arith- 
metic is  performed,  that  is  absolutely  necessary,  if  the  pupil  icill 
only  kee]^  the  ^:>ri?zc'ipZ6'5  of  numbers  firmly  fixed  in  his  mind,  and 
without  this  he  cannot  succeed  if  he  has  a  book  full  of  forms.  They 
are  used  only  for  looks  and  convenience  in  illustrating  our  operations. 

It  will  be  noticed  that  each  figure  of  the  quotleiU  occupies  the  same 
place,  numerically,  as  the  figure  of  the  dividend  that  ivas  divided 
to  produce  it.     This  is  always  the   case. 


What  is  the  number  to  be  divided  called  ?  The  number  we  divide  by?  The 
result  or  answer?  From  what?  What  does  Quoties  signify?  What  is  the 
sign  of  Division?  What  does  it  show?  How  is  division  otherwise  ex- 
pressed? Give  the  solution  of  Ex.  1st.  Is  there  any  other  method  of  working 
such  questions?  How  do  we  write  the  numbers  by  this  method?  How  pro- 
ceed next?  Is  this  form  of  any  particular  importance?  Why  then  is  it  used? 
Is  any  form  absolutely  necessary  in  arithmetic?  What  must  a  scholar  do  if  he 
dispenses  with  forms?  Can  he  succeed  well  without  this?  Wiiy  not?  What 
then  is  the  use  of  forms?  What  place  does  each  figure  occupy  numerically? 
Is  this  alwayjj  the  case ?^ 


Art.  2.  SIMPLE   DIVISION".  89 

2.  A  man  spent  488  cents  for  ribon,  at  4  cents  a  yard  ;  how 
many  yards  did  he  get?  Ans.   122. 

3.  Four  men  received   845  dollars  ;  how  much  was  that  apiece? 

Ans.  211 5:  dollars. 

In  this  example,  after  dividing  by  4,  there  was  1  dollar  left. — 
Now  this  1  dollar  is  to  be  divided  between  4  men  ;  but  as  1  will  not 
contain  4,  we  will  merely  express  the  division  ;  thus  :  j.  This  is 
called  one-fourth,  and  annexing  it  to  the  right  of  our  quotient,  we 
find  that  each  man  received  211^  dollars. 

The  1  which  was  left  after  the  division  was  performed,  is  called 
a  REMAINDER.     Hcnce — 

Obs.  6.  When  there  is  a  remainder,  it  is  written  over  the  divi- 
sor, and  annexed  to  the  quotient. 

Obs.  7.  It  will  be  perceived  that  Division  is  exactly  the  reverse 
of  Multiplication.  In  the  latter  we  have  two  factors  given  to  find 
their  product :  in  the  former  we  have  the  product  and  one  factor 
given  to  find  the  other  factor,  the  quotient  and  divisor  answering  to 
to  the  two  factors,  and  the  dividend  answering  to  to  the  product,  in 
Multiplication.     Hence — 

Obs.  8.  If  we  multiply  tJie  divisor  and  quotient  together,  we 
shall  obtain  the  dividend. 

But  if,  after  a  division  has  been  performed,  there  is  a  remainder, 
it  is  evident  that  the  product  of  the  quotient  and  divisor  will  want 
just  this  remainder  to  make  the  dividend.     Hence — 

To  prove  Division  : 

Obs.  9.  Mdtiply  the  divisor  and  quotient  together,  and  to  the 
product  add  the  remainder,  {if  any)  ;  if  the  residt  is  equal  to  the 
dividend,  the  work  is  correct. 

From  the  preceding  remarks,  we  derive  the  following  method  of 
proving  Multiplication  by  Division  : 

Obs.  10.  Divide  the  product  hy  one  factor,  and  if  the  residt  s 
the  other  factor,  the  ivork  is  correct. 

NoTK. — In  this  case  there  is  never  any  remainder.* 

4.  At  3  dollars  a  yard,  how  many  yards  of  cloth  can  be  bought 
for  697  dollars?  _  Ans.  232^. 

.Let  the  learner  prove  this  and  the  following  examples  : 

6.  At  5  dollars  an  acr^-,  how  many  acres  of  land  can  be  bought 
for  568  dollars?  Ans.   111|. 

If  there  is  a  number  left  after  dividing^,  what  is  it  called?  What  is  done  with 
it'?  Of  what  is  Division  the  reverse?  Show  the  comparison.  To  what  do  the 
different  terms  of  Division  answer  in  Multiplication?  What  inference  is  de- 
duced from  this?  But  suppose  after  the  division  has  been  performed,  there  is 
a  remainder  ;  what  then  is  the  conclusion?  How  then  do  we  prove  Division? 
How  can  we  prove  Multiplication  by  Division?  Is  there  any  rejnainder  in 
this  case?     Why  not? 

•TftRtls,  In  Integra!  numUert. 


40  COMMON  ARrruMETio.  Sect.  V. 

6.  I  wish  to  set  out  l^G  trees  in  6  rows  ;  how  many  must  I  set  in 
a  row?  Ans.  21.. 

Ill  this  example  l-.  ,  ,.  ..  ::.^ure  of  the  dividend  did  no&  contain  the 
divisor,  therefore  we  look  hvo  iigUies  ;  if  two  figures  hud  not  eon- 
tained  it,  we  should  have  taken  three  figures,  &c.     Hence — 

Obs.  11.  If  the  left  hand  figure  of  the  dividend  will  not  con- 
tain the  divisor,  take  two  figures  ;  if  two  figures  will  not  contain  it, 
take  three  figures ;  and  universally  take  as  many  figures  at  the  left 
of  the  dividend,  as  ivill  contain  the  divisor  once  or  more,  to  obtain 
the  first  quotient  figure,  which  we  write  under  the  right  hand  figure 
of  the  part  taken. 

7.  Six  men  had  366  dollars  to  divide  between  them.  How  much 
was  that  apiece?  Ans.  61  dollars. 

8.  A  man  had  976  acres  of  land,  which  he  wished  to  divide  be- 
tween his  4  sons.     How  many  acres  would  each  receive? 

Operation.  976  =  9   (hundreds)  +  7  (tens)  +6.      9 

4)976  (hundreds)   =  8  (hundreds)  -f-  1  (hundred). 

Now  4  will  go  in  8( hundreds)  just  2(  hundreds) 

Ans.  244  acres.  times  ;  consequently  it  will  go  in  9  (hundreds) 
2  (hundreds)  times,  and  have  1  (hundreds)  remainder.  Adding  the 
1  (hundreds)  remainder,  (which  is  10  tens,)  to  the  7  (tens),  we  have 
17  (tens,)  which  divided  by  4  gives  4  (tens)  for  a  quotient,  and  1 
(ten)  remainder.  This  1  (ten)  is  equal  to  10  imits,  which  added 
to  6  (^units)  gives  16  (units)  This  divided  by  4  gives  4  (units),  and 
our  work  is  done.  Hence,  onr  quotient  is  2  (hundreds),  4  (tens), 
and  4  (units),  or  244. 

Obs.  12.  By  attentively  examining  the  above  solution,  the 
learner  will  perciive  the  following  considerations,  viz  : 

a.  1st.  The  remainder  each  time  is  of  the  same  name  as  the  div- 
idend. T/iis  is  always  the  case,  whether  the  remainder  occurs  af-. 
ter  each  quotient  figure  is  obtained,  or  only  at  the  end  of  the  opera- 
tion ;  because  the  remainder  is  always  a  part  of  the  dividend. 

b.  2nd.  The  remainder  each  time  should  be  less  than  the  divi- 
sor;  because,  if  it  Y/ere  eq^lal  'to,  or  greater  than  the  divisor,  the  di- 
visor could  be  contained  another  time  in  the  dividend. 

c.  3d.  The  remainder  each  time  has  ten  times  the  value  of  the 
next  lower  order.  This  is  evident,  according  to  Sect.  1,  Art. -2, 
Obs.  3.     Hence,  we  must  multiply  it  by  10  before  adding  it  to  the 

If  the  first  figure  of  the  dividend  will  not  contain  the  divisor,  hovi^  do  we 
proceed?  How  iriany  figures  should  we  tak*^,  in  ail  cases  from  the  dividend, 
to  obtain  our  first  quotient  figure?  Explain  the  sohuion  of  Ex.  8.  What  is 
the  first  consideration  deduced  from  ihis?  Is  this  always  the  case?  Why? 
What  is  the  second  consideration?  Wijy?  What  is  the  third  consideration? 
Why?  What  then  musl  be  done  before  it  can  be  added  to  the  next  lower 
figure? 


Art.  2.  SIMPLE  uivisioiV.  41 

next  figure,  (Sect.  II.,  Art,  2,  Obs.  4.)  It  has  the  same  effect, 
however,  to  prefix  it  to  the  next  figure,  as  the  cipher  always  occu- 
pies the  place  filled  by  this  figure. 

Note. — Prefix  means  to  place  before,  or  at  the  l^t  hand. 
This  may  be  illustrated  as  follows  : 
0.   Divide  1074  by  3. 

Operation.  In   the  first    case    we   have    1     remainder, 

3)10     7     4      which,  multiplied   by    10,  equals    10,    and    7 

10  20      equals  17.     We  next  have  2  remainder,  which, 

multiplied  by  10,  equals   20,  and  4  makes  24, 

1 7  24      which  contains  3  without  a  remainder. 

Hence — 

3     5     8 
Obs.   13.      When  there  is  a  remainder  after  dividing  any  figure 
of  the  dividend,  prefix   it  {mentally)  to  the  next  figure  of  the  divi- 
dend, and  proceed  as  before. 

10.  At  8  dollars  an  acre,  how  many  acres  of  land  can  be  bought 
for  8984  dollars?  Ans.   1123. 

11.  At  5  dollars  a  pair,  how  many  pair  of  boots  may  be  bought 
for  685  dollars?  Ans.   137. 

12.  At  4  shillings  a  bushel,  how  many  bushels  of  potatoes  can 
be  obtained  for  1628  shillings. 

Operation.  In  this  case  the  divisor  is  not  contained  in  the 

4)1628  ten's  figure,  therefore  we  write  a  cipher  in  the 

quotient,  and  prefix  the  2   to  the  next  figure,  as 

407  if  it  were  a  remainder,  which  it  really  is.     Then 

4  in  28  goes  7  times.     Hence — 

Obs.  14.  When  any  partial  dividend  loill  not  contain  the  divi- 
sor, write  a  cipher  in  the  quotient,  call  the  p)artial  dividend  a  re- 
mainder, and  proceed  as  before. 

Remark. — The  partial  dividend  is  tliat  part  taken  to  find  each  quotient  fig- 
ure. 

Obs.  15.  Also — After  the  first  quotient  figure  is  obtained,  either 
a  cipher  or  a  significant  figure  must  be  pid  in  the  quotient,  for  each 
figure  or  cipher  in  the  dividend. 

13.  A  man  traveled  1624  miles  in  8  weeks;  how  many  miles 
was  that  a  week?  ^  Ans.  203. 

Why?  By  what  other  method  can  we  effect  the  same  result?  Why  is  this 
correct?  What  does  prefix  mean?  Give  the  solution  of  Ex.  9,  and  show  why 
it  has  the  same  effect  to  prefix  the  remainder  to  the  next  lower  figure  as  to 
multiply  it  by  10,  and  then  add  it.  What  inference  is  deduced  from  (his? — 
When  the  divisor  cannot  be  contained  in  any  partial  dividend,  how  do  we  pro- 
ceed? What  is  the  partial  dividend?  How  many  figures,  or  cipher.-:,  must  be 
put  in  the  quotient  after  the  first  quotient  figure  is  obtained?  When  the  divi- 
sor is  small,  and  the  operation  is  carried  on  chiefly  in  the  mind,  what  is  it 
called? 


4^'  COMMON    ARITHMETIC.  Sect.    V. 

When  the  divisor  is  small,  as  in  the  preceding  examples,  the  ope- 
ration is  performed  chiefly  in  the  mind  ;  it  is  then  called  Short  Di- 
vision. 

From  the  remarks  and  illttstrations  already  given,  "we  derive  the 
following 

RULE  FOR  SHORT  DIVISION. 

I .  Write  ike  divisot'  at  ike  left  of  ike  dividend,  ivitk  a  line  be- 
tween them;  also,  draw  a  line  below  ike  dividend, 

II.  Take  from  ike  left  of  ike  dividend  as  many  figures  as  ivill 
contain  ike  divisor  once  or  more;  divide  ikon,  write  ike  quotient 
under  ike  rigki  kand  figure  of  ike  part  taken,  (Obs.  11)  /  prefix 
the  remainder  [if  any,)  to  ike  next  figure  of  ike  dividend,  and  di- 
vide as  before,  uyUil  all  ike  figures  of  ike  dividend  kave  been  di- 
vided. (Obs.  13.) 

III.  If  the  divisor  is  not  contained  in  any  partial  dividend^ 
ivrite  a  cipker  in  the  quotient,  prefix  this  partial  dividend  to  the  next 

figure,  as  if  it  were  a  remainder,  and  proceed  as  usual.  (Obs.  14.) 
ly .     If  a  remainder  occurs  after  dividing  ike  last  figure  of  ike 

dividend,  write  it  over  ike   divisor,  and  annex  it  to  ike  quotient. — 

(Obs.  6.) 

Proof. — Mkdtipily  ike   divisor  and   quotient    togetker,    and  to  ike 

product   add  ike   remainder  {if  any)  ;    if  the  sum   is  equal  to  ike 

dividend,  ike  ivorTc  is  correct.  (Obs.  9.) 

1.  We  commence  at  the  left  hand  to  perform  division,  because 
the  remainders  which  occur  must  be  reduced  to  lower  orders  before 
the  division  can  be  continued.  (Obs.  12,  3d.)  Besides,  it  is  evi- 
dent that  if  we  divide  any  order  ot  figures  by  any  number,  and 
there  is  a  remainder,  that  tkis  remainder  cannot  be  reduced  to  a 
kigker  order,  and  still  contain  ike  same  divisor.  Hence,  it  must 
be  reduced  lower;  and  as  numbers  decrease  from  the  left  hand 
towards  the  right,  we  commence  at  the  left  to  divide.  The  learner 
will  better  understand  the  reasons  for  carrying  in  the  Fundamental 
Rules,  v/hea  he  has  become  acquainted  with  Compound  numbers. 

2.  The  divisor^  dividend  and  quotievt  may  any  of  them  be  either 
abstract  or  concrete.  The  divisor  and  quotient,  however,  cannot 
both  be  concrete  in  the  same  question,  and  when  eitker  of  them  is 
concrete,  the  dividend  is  also  concrete,  and  of  the  same  name.     Nei- 

What  is  the  rule  for  Short  Division?  Why  do  we  commence  at  the  left 
hand  to  divide?  Why  must  the  remainders  be  reduced  to  lower  orders  before 
we  can  continue  the  operation?  What  is  said  respecting  the  divisor,  dividend, 
and  quotient  being  abstract  or  concrete?  Can  the  divisorand  quotient  botii  be 
concrete  in  the  same  question?  When  one  of  them  is  concrete,  what  must  tho 
dividend  be?  Can  the  dividend  be  abstract,  without  the  divisor  and  quotient 
both  being  abstract  ?  What  name,  tlien,  must  the  dividend  always  be?  Give 
examples. 


Art.   2.  SIMPLE   DIVISION.  4S 

tlier  can  the  dividend  be  abstract,  without  the  divisor  and  quotient 
both  being  abstract  also.  In  short,  the  dividend  must  alwai/s  be  of 
the  same  name  as  the  divisor,  or  quotient,  or  both.  Thus  :  in  the 
6th  question,  we  reason — If  126  trees  are  put  in  6  rows,  each  row 
will  contain  one-sixth  of  126  trees,  which  is  21  trees,  and  not  21 
times,  or  merely  21,  as  is  often  asserted.  Here,  the  divisor  (6)  is 
abstract,  whilst  the  dividend  and  quotient  are  both  concrete,  and  of 
the  same  kind.  In  the  11th  example  we  divide  685  dollars  by  5 
dollars;  but  ihe  learner  must  not  infer  that  685  dollars  divided  by  5 
dollars  gives  137  boots  for  an  answer.  In  this  case,  as  the  boots  cost 
5  dollars  a  pair,  we  can  buy  as  many  pair  as  5  dollars  is  contained  in 
685  dollars,  which  is  137  times.  Here  our  divisor  and.  dividend  are 
concrefe,  and  of  the  same  name,  and  the  quotient  is  abstract.  It  would 
be  as  absurd  to  make  the  quotient  concrete  in  this  case,  as  it  would 
be  to  say,  (in  the  8th  Ex.)  that  976  acres  contain  4  sons  244  times, 
or  that  one  fourth  of  976  acres  is  any  ihing  else  than  244  acres. 

EXERCISES    FOR    THE    SLATE. 

1.  At  6  cents  a  ball,  how  many  balls  of  tape  may  be  bought  for 
1^6  cents?  '  Ans.  31. 

2.  How  many  sheep  can  I  buy  for  786  dollars,  at  3 dollars  aj^iece? 

Ans.  262. 

3.  How  many  pair  of  gloves  can  I  buy  for  1728  shillings,  at  9 
shillings  a  pair?  Ans.   192. 

4.  There  are  3  feet  in  a  yard  ;  how  many  yards  in  479  feet? 

Ans.   159|. 

5.  Amai  bought  9  horses   for  1382  dollars.     How  much  was 
that  apiece?  Ans.   153f  dollars. 

6.  A  man  sold  6  yoke  of  oxen  for  492  dollars.     How  much  was 
that  a  yoke?  "  Ans.  82  dollars. 

7.  If  7  pieces  of  cloth  cost  438  dollars,  how  much  is  that  apiece? 

Ans.  624  dollars. 

8.  If  8  wagons  cost  548  dollars  what  is  that  apiece? 

Ans.  68|-  dollars. 

9.  At  4  shillings  apiece,  how  many  books  can  be  bought  for  6488 
shillings?  Ans.    1622. 

10.  At  11  miles  a  day,  how  many  days  would  it  take  to  travel 
483  miles?  x\ns.  431-t- 

1 1 .  There  are  8  quarts  in  a  peck;  how  many  pecks  in  9952  quarts? 

Ans.   1244. 

12.  At  5  cents  a  mile,  how  far  can  I  ride  for  8730  cents? 

Ans.   1746  miles. 

13.  At  12  dollars  a  month,  how  many  months  would  it  t^ke  to 
earn  4623  dollars?  Ans.  385y\. 

14.  How  many  calves,  at  7  dollars  apiece,  can  I  buy  for  3G8  dol- 
laj-s?     .    (u  . -.  .,.   -.  .  Ans.  52,  ^y.d  have  4. dollars  left,   . 


44  COMMON    ARITHMETIC.  Scct.     V. 

15.  A  man  has  2832  pair  of  shoes  in  8  boxes.  How  many  are 
there  in  each  box?  Ans.  354  pair. 

16.  At  5  dolhirs  a  barrel,  how  many  barrels  of  flour  can  I  buy 
for  2540  dollars?  Ans.   508. 

17.  If  8  laborers  should  perform  a  piece  of  work  for  1628  dol- 
lars, how  much  would  that  be  apiece?  Ans.  203|-  dollars. 

18.  16  tmx'S  489271  is  how  many  times  2?       Ans.  3914168. 

19.  15  limes  14962  is  how  many  times  5?  Ans.  44886. 

20.  17  times  6217  is  how  many 'times  11?  Ans.  9608yV 

Article  3.     Long  Division. 

Obs.  1 .  In  Short  Division  the  operation  is  performed  mentally  to 
a  considerable  extent ;  but  in  Long  Division  the  operation  is  all  set 
down  upon  the  slate.  The  principles  of  both,  however,  are  the 
same.  H^C;  former  is  gene'  ally  used  when  the  divisor  is  less  than  12, 
and  the  latter  when  the  divisor  exceeds  12. 

1.  A  man  paid  4096  dollars  for  some  land,  at  16  dollars  an  acre. 
How  many  acres  did  he  buy? 

Operation.. 

We  write  our  numbers  as  before,  except  16)4096(256  Anst 
the  quotient,  which  we  place  at  the  right  32  •  • 

hand.     We   do  this  lecause  it  is  more  con-  

venient  than  to  place  it  beloiv.     We  next  say,  89 

16  in   40  (hundreds)    goes  2    (hundreds)  80 

times,  and  write  our  2  in  the  quotient.     We  

then  multiply  our  divisor  (16)  by  our  quo-  96 

tient  figure  (2),  and  place  the  product  (32)  96 

under   the  part  taken   of  the  dividend. —  — 

This  is  done  to   ascertain    whether  there  is  00 

any  remainder  after  dividing.  Next  we  subtract  this  product  (32) 
from  the  dividend,  (40,)  which  leaves  8(hundreds)  as  our  remainder. 
We  then  bring  down  the  9  (tens)  at  the  right  of  the  8  (hundreds,) 
(which  is  the  same  as  prefixing  the  remainder  to  the  9  (tens),  which 
makes  89  (tens).  Then  16  in  89  (tens)  goes  5  (tens)  times,  which 
we  write  in  the  quotient.  We  place  the  5  at  the  right  of  the  2 
lecause  the  2  is  himdreds,  and  the  5  is  tens,  and  tens  occupy  the 
place  at  the  right  of  hundreds.  Then  5  times  16  is  80,  which  we 
place  under  the  89,  and  subtracting,  we  have  9  (tens)  as  our  second 
remainder.  To  this  we  bring  down  our  next  figure  in  the  divi- 
dend (6),  making  96  (units),  which  divided  by   16,  gives  6  (units) 

What  is  the  difference  between  the  operation  of  Long  and  Short  Division? 
What  is  the  difference  between  the  principles  of  the  two?  When  is  Short  Di- 
vision generally  used?  Long  Division?  How  do  we  write  the  numbers  in 
Long  Division?  Why  do  we  place  the  quotient  on  the  right?  What  is  the 
first  step  of  the  operation?  The  second?  Third?  Fourth?  Why  do  we 
multiply  the  divisor  by  the  quotient  figure?  What  is  meant  by  bringing - 
down?     Why  do  we  place  our  second  quotiout  figure  at  the  right  of  the  first? 


Art.  3.  LOXG  Div.sioN  45 

for  our  next  quotient  figure.  Finally,  multiplying  and  subtracting 
as  before,  we  find  there  is  no  remainder ;  and  as  all  the  figures 
of  the  dividend  have  been  brought  down  and  divided,  our  work  is 
done. 

Note. — The  dots  are  placed  under  figures  of  the  dividend  when  they  are 
brought  down,  to  prevent  their  being  brought  down  again.  They  are  not  neces- 
sary if  the  pupil  is  observing. 

Obs.  2.  By  examining  this  solution  attentively,  the  learner  will 
observe  the  two  following  considerations,  viz  : 

1st.  £ack  fiffure  in  the  quotient  occupies  ike  same  2J>Iace,  numeri- 
cally, as  the  right  hand  figure  of  that  part  of  the  dividend  taken  to 
produce  it.  (Art.  2,  Obs.  5.) 

2nd.  The  operations  of  Long  Division  consist  of  four  stepsy 
viz:  1st,  to  divide;  2nd,  to  midtiply ;  3rd,  to  subtract;  and  4  h, 
to  bring  down.     This  is  the  whole  process. 

2.  A  man  bought  15  acres  of  land  for  375  dollars.  How  much 
was  that  per  acre?  Ans.  25  dollars. 

3.  If  a  man  travel  625  miles  in  25  days,  how  many  miles  is  that 
per  day?  Ans.  25. 

4.  A  man  paid  32  dollars  an  acre  for  some  land,  and  bought 
enough  to  come  to  34576  dollars.     How  many  acres  did  he  buy? 

Operation. 

32)34376(1080^1.         32  in  34  goes  1  time,  and  2  remainder. 

32  Bring  down   the   5,  we  find  thai  25  will 

—  not  conta  n  32  ;  therefore  we  write  a  ci- 

257  pher  in  the  quotient,  (Art.  2,  Obs.  14), 

2'^^  and  bring  down  the  next  figure,  7.     32 

in  257  goes   7  times,  and   1  remainder. 

16  Bringing  down   the  6,  we  find  that  16 

will  not  contain  the  32,  and  therefore  write  another  cipher  in  the 
quo  lent ;  then  as  there  are  no  more  figures  in  the  dividend  to  bring 
down,  16  is  our  remainder,  which  we  write  over  our  divisor,  and 
annex  to  the  quotient,  (Art.  2,  Obs.  6),  and  thus  obtain  for  our 

Ans.   10801J. 

5.  How  many  times  will  6144  contain  24? 

1st  Operation. 
dh.i  24)6 144(257 
48 

—  Here,  we  supposed  our  last  quotient  figure 
134  to  be  7  ;  but  by  multiplying,  we  find  that  our 
120                    product  is  greater  than  our  dividend  ;  therefore 

the  quotient  figure  is  too  large. 

144 
168 


f^f 


40  COMM3N     ARITHMETIC.  Scct.    Y. 


2(1    Operation. 

24)6144(256 

48 

134 

Here,  we  supposed  it  to  be  6, 

■\vliicli  we 

120 

find  to  be  correct. 



Hence — 

144 

- 

144 

000 

Obs.  3.  When  the  product  of  the  divisor  liy  any  quotient  figure 
is  greater  than  the  dividend,  the  quotient  figure  must  be  made  less. 
The  rem  ^rks  respecting  the  remainder,  (x\rt.  2,  Obs.  12,)  ate  equ  .Uy 
as  applicable  to  Long  as  to  Short  Division. 

From  the  preceding  remarks  and  illustrations,  we  derive  the  fol- 
lowing 

RULE  FOR  LONG  DIVISION. 

I.  Write  tJie  divisor  and  dividend  as  before  directed,  and  drdio 
a  line  at  the  right  of  the  dividend  to  separate  it  from  the  quotient. 

II.  Take  as  many  figures  at  the  left  of  the  dividend  as  vjill  con- 
tain the  divisor  once  or  more,  and  place  the  number  of  times  they 
contain  it  at  the  right  of  the  first  quotient  figure. 

in.  Multiply  the  divisor  by  this  quotient  figure,  and  subtract  the 
product  from  that  part  of  the  dividend,  taken. 

IV.  To  the  right  of  the  remainder  bring  down  the  next  figure 
of  the  dividend,  and  proceed  as  before.  So  continue  to  do  till  all 
the  figures  of  the  dividend  have  been  brought  down  and  divided. 

V.  If  any  partial  dividend  will  not  contain  the  divisor,  write  a 
cipher  in  the  quotient,  bring  down  the  next  figure  of  the  dividend, 
and  proceed  as  before.  (Art.  2,  Obs.  14.) 

Proof. — Multiply  together  the_  .divisor  and  quotient,  and  to  the 
product  add  the  remaiTider  {if  any)  ;  if  the  result  is  equal  to  the 
dividend,  the  work  is  correct.  (Art.  2,  Obs.  9.) 

EXERCISES    FOR    THE    SLATE. 

1.  If  a  man  walk  4  miles  an  hour,  how  many  hours  will  it  take 
him  to  walk  123  miles?  Ans.  32. 


For  what  are  the  dots  used?  Are  they  really  necessary?  Wliat  place 
does  each  quotient  figure  occupy  nurnericaily?  Of  how  many  steps  does 
the  operation  in  Long  Division  consist?  Name  them.  When  the  dividend 
will  not  contain  the  divisor,  how  do  we  proceed?  What  is  done  vt-ith  Ihe  final 
remainder?  When  the  product  of  the  divisor  by  any  quotient  figure  its  greater 
than  the  dividend,  what  is  to  be  done?  Are  the  remarks  respecting  the  re- 
mainder, (Art.  2,  Obs.  8.)  applicable  to  Long  as  we'd  as  Short  Division?  What 
wore  theio  remarks?    What  is  the  rule  for  Long  Divi^ioji? 


Art.  3,  LONG   DIVISION.  47 

2.  There  are  12  months  in  a  year  ;  how  many  years  in   20176 
months?  Ans.   1681 -j. 

3.  There  are   8  quarts  in  a  peck.     How  many  pecks  in  32024 
quarts?  Ans.  4003. 

.4  There  are  6  working  days  in  a  week.  How  many  weeks  in 
1256  working  days?  Ans.  209|. 

5.  A  half  eagle  is  worth  5  dollars.  How  many  half  eagles  would 
it  take  to  be  worth  394820  dollars?  "      Ans.  78964. 

6.  In  one  square  yard  are  9  square  feet.  How  many  square 
yards  in  42786  square  feet?  Ans.  4754. 

^-  7.  How  many  calves,  at  14  dollars  apiece,  can  be  bought  fur  476 
dollars?  Aiis.  34. 

8.  At  17  dollars  a  month,  how  many  months  would  it  take  to 

.earn  969  dollars?  Ans.  57. 

her  9.  The  steamer  Great  Western  was  15  days  in  crossing  the  At- 
lantic Ocean,  a  d'stance  of  3000  miles.  How  far  did  she  sail  per 
day?  r-Y^A-ryAnryfy?  ^^^'  ^^^  miles. 

10.  Suppose  21528  dollars  were  equally  divided  between  104 
men;  how  much  would  each  receive?  Ans.  207  dollars. 

11.  Suppose  a  vessel  has  sailed  15340  miles,  at  the  rate  of  126 
miles  per  day  ;  how  many  days  has  she  sailed?      Ans.    121  j/g-- 

12.  There  are  63  galloiis  in  a  hogshead.  How  many  gallons  in 
7560  hogsheads?  Ans.   120. 

13.  At  16  dollars  a  month,  how  long  would  it  take  to  earn  624 
dollars?  Ans.  39  months. 

14.  How  m  ny  days  would  it  take  to  walk  576  miles,  walking  24 
miles  per  day?  Ans.  24  days. 

15.  In  every  day  are  24  hours.     How  many  days  in  4096  hours? 

^  .  Ans.   17011. 

16.  At  18  dollars  a  ton,  how  many  tons  of  hay  could  be  bought 
for  450  dollars?  "  Ans.  25. 

17.  How  many  casks  of  wine  could  be  bought  for  4950  dollars, 
at  25  dollars  a  cask?  Ans.   198. . 

18.  If  a  man  has  traveled  6255  miles  in  15  weeks,  how  far  is 
that  per  week?  Ans.  417  miles. 

19.  If  75  wagons  cost  5625  dollars,  how  much  is  that  apiece? 

Ans.   75dolla's. 

20.  If  53  horses  cost  3286  dollar?,  how  much  is  that  for  eacb?. 

.  '_  ...  Ans.  62  dollars. 

21.  If  32  pounds  of  raisins  cost  800  cents,  how  much  is  that  per 
pound?  Ans.  25cens. 

22.  A  man  has  a  note  to  pay,  amounting  to  1204  dollars.  If  he 
pays  86  dollars  at  a  time,  how  many  payments  will  he  have  to 
make  to  pay  up  his  note?  Ans.   14. 

23.  There  are  62  weeks  in  a  year.  How  many  years  in  1248 
weeks?  Ans.   24. 


48  COMMON    ARITHMETIC.  Scct.    V. 

24.  Suppose  a  firm  to  consist  of  37  partners,  and  their  capital  to 
amount  !o  1178450  dollars.     How  much  is  that  for  each? 

Ans.    31850  dollars. 

25.  At  144  shillings  a  month,  how  long  would  it  take  a  man  to 
earn  7488  shillings?  Ans.  52  months. 

26.  25  times  8469  is  how  many  times  9?  Ans.  23525. 

27.  144  Times  7324  is  how  many  times  48?  Ans.  21972. 

28.  1728  times  20736  is  how  many  times  288?    Ans.    124416. 
2).  Two  numbers  multiplied  together  produce  one  billion,  thirty 

million,  six  hundred  and  three  thousand,  six  hundred  and  fifteen, 
imd  one  of  the  numbers  is  three  thousand,  two  hundred  and  fifteen. 
What  is  the  other  number?  Ans.   32056 1 . 

30.  Divide  th  rty-seven  billion,  four  hundred  and  twenty -three 
million,  ei-ht  hundred  ;  nd  :hirty-four  thousand,  five  hundred  and 
sixty,  by  one  hundred  and  twenty-three  thousand,  four  hundred  and 
fifty-six.  *  Ans.   303135. 

Article  4.     Contractions  in  Division. 

Case  l.—To  divide  hy  10,  100,  1000,  &c. 

1.  At  10  dollars  a  month,  how  many  months  would  it  take  to 
earn  534  dollars? 

Operation,  As  a  cipher  annexed  to  any  figure  increases 

1|0  I  53|4        its  value   10  times,   (Sect.   1,    Art.    1,  Obs.   5,) 

and  two  ciphers  annexed  increases  its  value  100 

Ans.  53yo  •  times,  &c..  so  to  cut  off  one  figure  from  the  right 
of  any  number,  diminishes  the  value  of  that  number  10  times,  to 
cut  oil"  two  figures  dimiiiishes  its  value  100  times,  to  cut  oflf  three 
fiofures  diminishes  its  value  1000  times,  &c. 

Hence — To  divide  by  1  with  any  number  of  ciphers  annexed  : 
Obs.  1.      Out  off  from  the  right  of  the  dividend  as  many  figures  as 
there  are  ciphers  at  the  right  of  the  divisor ;  the  figures  at  the  left 
will  he  the  quotient,  and  those  at  the  right  the  remainder. 

2.  Suppose  you  wished  to  divide  4267  dollars  equally  between 
100  men,  how  much  would  you  give  to  each  man? 

Ans.  42yVo  dollars. 

3.  There  are  10  mills  in  a  cent.     How  many  cents  in  150  mills? 

4.  There  are  100  cents  in  a  dollar.  How  many  dollars  in  4875 
cents? 

5.  There  are  10  dollars  in  an  eagle.  How  many  eagles  in  3854 
dollars  ? 

6.  There  are  1000  mills  in  a  dollar.  How  many  dollars  in  65832 
mills? 

7.  There  are  10000  mills  in  an  eagle.  How  many  eagles  in 
489621  mills? 

How  do  we  divide  by  10, 100, 1000,  &c.?    Explain  why  thia  rule  is  correct. 


Art.    4.  CONTRACTIONS   IN    DIVISION.  49 

Case  ^     To  divide  hy  a  Composite  Number, 

If  we  divide  12  by  6,  we  obtain  2.  Again  :  if  we  divide  12  by 
3,  we  obtain  4  ;  and  if  we  divide  4  by  2,  we  also  obtain  2.  Now 
3  ;md  2,  (our  litter  divisors,)  are  the  factors  of  6,  and  to  divide  12 
by  them  produces  the  same  result  as  to  divide  12  by  6.     Hence — 

Obs.  2.  To  divide  the  dividend  by  the  factors  of  the  divisor ,  will 
produce  the  same  result  as  to  divide  it  by  the  divisor  itself. 

a.  Therefore,  conversely,  Any  dividend  that  will  contain  a  divisor 
will  also  contain  the  factors  of  that  divisor. 

1 .  If  I  pay  240  dollars  for  48  yards  of  cloth,  what  is  that  per 
yard?  Ans.  5  dollari. 

1st  Operation.  2d  Operation. 

48)240(5  dollars.    Ans.      8)240 

240  

6)30 

000  — 

Ans.  5  dollars. 
48  is  composed  of  two  factors,   6  and   8,  (8  X  6  =  48,)  and  ac- 
cording to  the  above  remark,  we  may  divide  240  first  by  one  fac- 
tor, and  t'le  quotient  thence  arising  by  the  other,  and  it  will  produce 
the  same  result  as  to  divide  it  by  48,  that  is,  5  dollars  a  yard. 
Hence — When  the  divisor  is  a  composite  number  : 
Obs.  3.     Divide  the  divide^id  frsi  by  one  factor,  and  the  quotient 
thence  arising  by  another  factor,   and  so  on  ;  the  last  quotient  will  be 
the  answer. 

Note. — It  makes  no  difference  in  any  case  which  factor  we  diride  by  first; 
if  no  mistake  is  made,  the  final  result  will  be  the  same. 

2.  If  a  man  travel  32  miles  per  day,  how  many  days  will  it  take 
him  lo  travel  1472  miles?     (32  =  4  X  8.)  Ans.  46. 

3.  If  a  man  pay  2835  dollars  for  horses,  at  the  rate  of  63  dollars 
apiece,  how  many  can  he  buy?     (63  =  9  X  7.)  Ans.  45. 

4.  How  many  acres  of  land,  at  25  dollars  an  acre,  can  be  bought 
for  875  dollars?     (25  =  5  X  5.)  Ans.  35. 

5.  How  many  yoke  of  oxen  can  I  buy  for  1904  dollars,  at  6Q  dol- 
lars a  yoke?     (56  =  8  X  7.)  Ans.  34. 

6.  How  many  watches  can  be  bought  for  1875  dollars,  at  75  dol- 
lars apiece.?     (75  =  5  X  5  X  3.)  Ans.  25. 

What  is  the  difference  in  the  result  of  dividing  12  by  6,  or  by  3  and  2,  the 
factors  of  6?  What  is  the  first  inference  deduced  from  this?  The  second? 
Give  the  solution  of  example  1st,  and  show  why  it  is  correct.  How  do  we  pro« 
ceed  when  the  divisor  is  a  composite  number?  Does  it  make  any  difference 
in  the  final  result  which  factor  we  divide  by  first?  Give  the  solution  of  Ex.  I, 
Case  3,  and  show  why  it  is  correct.  What  is  done  with  the  remainder,  3,  in 
Ex.  3?  How  do  we  proceed  when  there  are  ciphers  at  the  right  of  the  divieer? 
4 


^  COMMON    ARITHMETIC.  ScCt.  V. 

7.  A  school  teacher  wishes  to  divide  256  apples  between  32  pu- 
pils. HoAv  many  must  he  give  to  each?  (32  =  4X8.)     Ans.  8. 

8.  If  175  watches  cost  4375  dollars,  how  much  is  that  apiecel 
(175  =  6  X  5  X  7.)  Ans.  25  dollars. 

Case  3.      When  there  are  ciphers  at  the  right  of  the  divisor. 

1.  There  are  40  rods  in  a  furlong.     How  many  furlongs  in  689 
rods? 

1st  Operation.  -  2d  Operation.  * 

40)689(17^^.  Ans.  4|0)68|9 

40  —  :, 

Ans.   17^°^. 

289 

280  40  is  composed  of  two  factors,   4  and  10, 

—  (10  X  4  =  40) ;  therefore,  we  may  first  di- 
9  vide  by  ten,  (which  is  only  to  cut  off  one  fig- 
ure at  the  right — Obs.  1,)  and  then  divide  the  quotient  thence  ari- 
sing by  4,  (Obs.  3;)  this  is  done  by  cutting  off  the  cipher  at  the 
right  of  the  divisor,  and  a  figure  at  the  right  of  the  dividend,  and 
then  dividing  the  remaining  figures  of  the  dividen^i  (68,)  by  the 
remaining  figure  of  the  divisor  (4)  ;  the  figure  that  was  cut  off  from 
the  dividend  is  the  remainder. 

2.  There  are  60  minutes  in  an  hour.     How  many  hours  in  754 
minutes? 

Operation. 

6  0)75|4  In  this  example,  after  dividing  by  6,  there  was 

—  3  remainder,  to  which  we  annexed  the  4  that 

Ans.   12fJ  was  cutoff,  making 34  as  our  true  remainder. 

Hence — When  there  are  ciphers  at  the  right  of  the  divisor  : 
Obs.  4.  Cut  off  these  ciphers  ;  also,  cut  off  as  many  places  at  the 
right  of  the  dividend  ;  then  divide  the  remaining  figures  of  the  dividend 
hy  the  remaining  figures  of  the  divisor,  and  to  the  last  remainder  an- 
nex the  figures,  or  ciphers,  that  were  cut  off  from  the  dividend.,  for  the 
true  remainder. 

3.  How  many  horses  can  be  bought  for  6400  dollars,  at  80  dollars 
apiece?  Ans.  80. 

4.  At  30  dollars  an  acre,  how  much  land  can  be  bought  for  1568 
dollars?    ,  Ans.  52—  acres. 

5.  How  many  barrels  will  9700  pounds  of  pork  pack,  allowing 
200  pounds  to  the  barrel?  ^       Ans.  A^~~. 

6.  How  many  bins  would  it  take  to  contain  43200  bushels  of 
t^^heat,  allowing  each  bin  to  contain  320  bushels?  Ans.   135 

7.  Allowing  each  bin  to  contain  360  bushels,  how  many  would  it 
take?  Ans.  120. 


•^  RECAPITULATION.^  51 

8.  At  60  dollars  a  yoke,  how  many  yoke  of  oxen  could  be  bought 
for  480  dollars?  Ans.  8, 

9.  At  80  dollars  a  yoke,  how  many  yoke  could  be  bought? 

Ans.  6- 

10.  An    army  of    13500   men    having   plundered  a  city,   took 
5130000  dollars.     What  was  each  man's  share? 

Ans,  380  d<?llars. 


SECTION  VL 
RECAPITULATIOX, 

Article  1.     Remarks  and  Inferences  deduced  from  the  Pre- 
ceding Rules. 

Obs.  1.  The  preceding  Rules,  viz. — Notation  and  Numeration 
Addition,  Subtraction,  Multiplication  and  Division — are  called  the 
Fundamental  Rules  of  Arithmetic,  because  they  are  the  basis,  or 
foundation  of  a  1  the  other  Rules,  and  without  the  aid  of  one  or 
more  of  these,  no  operation  in  numbers  can  be  performed. 

In  this  Section  it  is  intended  to  review  these  Rules,  and  make  a 
.  few  remarks,  and  present  a  few  ideas  and  suggestions  in  addition  to 
what  has  already  been  said.  The  rules,  as  a  general  thing  are  so 
evident,  that  they  need  no  demonstration.  If  the  pupil  cannot  de- 
monstrate them  himself,  he  should  turn  back  and  study  over  the 
preceding  Sections  again.  It  is  absolutely  necessary  that  he  should 
understand  the  preceding  Rules  thoroughly  before  he  can  ever  be- 
come an  arithmetician - 

Obs.  2.  Notation. — All  numbers  are  expressed  (by  tlie  common, 
or  Arabic  method,)  by  nine  digits  and  a  cipher.  As  the  cipher, 
however,  is  only  used  to  give  value  to  significant  figures,  (it  having 
no  value  itself,)  it  might  with  propriety  be  called  an  auxiliary  digit, 

Obs.  3.  Our  system  of  Notation  is  also  called  the  decimal  system, 
because  the  numbers  increxise  in  a  ten- fold  ratio.  The  term  dechncd 
is  derived  from  the  Latin  word  decern,  which  signifies  ten. 

Obs.  4.     The  names,  as  well  as  the  characters  which  represent 

What  are  the  preceding  rules  called?  Why?  Is  it  necessary  for  the  learner 
to  understand  these  rules,  before  he  can  become  an  arithmetician?  Why  «o? 
How  are  numbers  expressed  by  the  common  method  of  notation?  For  what  is 
the  cipher  used?  What,  then,  might  it  properly  be  called?  By  what  other  name 
is  the  common  system  of  notation  known?  Why  is  it  thus  called?  From 
what  is  the  term  decimal  derived?  What  is  said  respecting  the  names  and 
characters  that  we  use  to  represent  numbers?  Would  an^y  other  names  or  char- 
acters have  answered  as  well  had  they  been  adopted  T  .    .     -^ 


52  COMMON    ARITHMETIC.  Scct.  VI. 

numbers,  are  entirely  arbitrary.  Any  other  names  or  characters 
would  have  served  the  same  purpose  had  they  been  adopted.  It  is 
«^5e  only  that  renders  them  familiar. 

Obs.  5.  The  names  of  numbers  below  ten,  are  either  jjrimitive 
words,  or  derived  from  the  Latin.  Eleven  and  twelve  are  regarded 
by  some  as  derivative  words,  formed  from  ten  and  one,  ten  and  two  ; 
thus  :  eleven  means  ten  leave  one — that  is,  if  you  take  ten  from  it 
you  leave  one ;  twelve  means,  ten  leave  two — that  is,  if  you  take 
ten  from  it  you  leave  two.  This  is,  however,  rather  a  forced  signi- 
fication, and  they  are  generally  regarded  as  primitive  words. 

Names  of  numbers  higher  than  twelve  t;re  but  repetitions  of  the 
preceding  names.  Thus  :  thirteen,  fourteen,  (fee,  signify  three  and 
ten,  four  and  ten,  <fec.,  and  twenty,  tliirty,  &c.,  signify  two  tens,  three 
tens,  (fee,  and  so  on,  The  different  powers  of  ten,  as  one  hundred, 
one  thousand,  &c.,  are  primitive  words. 

Obs.  6.  Numeration. — It  may  be  asked  by  some,  why  do  toe 
coimnence  at  the  left  hand  to  read  7iumberSy  and  at  the  right  hand  to 
numerate  them?  The  reason  is  obvious.  In  speaking  of  numbers 
we  always  name  the  highest  orders  first,  as  it  is  more  convenient; 
and  therefore,  as  the  higher  orders  are  at  the  left  hand,  by  our  sys- 
tem or  notation,  we  commence  at  the  left  hand  to  read. 

Again — In  numerating,  we  call  the  lower  orders  first ;  therefore, 
as  numbers  increase  from  the  right  hand  towards  the  left,  the  lower 
orders  must  be  at  the  right ;  consequently,  we  commence  there  to 
numerate.  Had  our  system  of  notation  been  such,  that  numbers 
increased  from  the  left  hand  towards  the  right,  we  should  have  com- 
menced at  the  rigid  hand  to  read,  and  at  the  left  hand  to  numerate. 

Obs.  7.  Addition. —  We  cannot  add  numbers  of  different  names 
together  ;  because  if  I  have  5  pear  trees  and  8  cherry  trees  in  an 
orchard,  it  cannot  be  said  that  their  sum  is  \'i  pear  trees,  or  13  cJier- 
ry  trees.  We  can,  however,  say  that  their  sum  is  13  tree^.  Again: 
if  I  saw  5  hay  stacks,  7  wheat  stacks,  12  trees,  2  barns,  12  horses, 
and  15  cattle  in  a  lot,  their  sum  would  not  be  all  stacks,  trees, 
barns,  horses,  or  cattle  ;  but  if  asked  how  many  objects  I  siw  in  the 


Are  the  names  of  numbers  below  ten  primitive  or  derivative  words? 
What  are  eleven  and  twelve  regarded  by  some?  Formed  from  what? 
How?  Is  this  a  natural  signification?  How  are  they  generally  re- 
garded? How  are  numbers  higher  than  twelve  formed?  Give  exam- 
ples. What  is  said  respecting  the  different  powers  of  ten?  Why  do 
we  commence  at  the  left  liand  to  read  numbers?  Why  do  we  com- 
mence at  the  right  to  numerate?  Had  our  system  of  notation  been  such  that 
numbers  increased  from  the  left  hand  towards  the  right,  where  would  we 
have  commenced  to  read,  and  where  to  numerate?  Explain  why  we  cannot 
add  numbers  of  different  naioes  together.  What  is  necessary  in  adding 
sevei.il  lu.ialiLrs  iojjelher?  When  wo  iiave  feiveu  seve-ral  numbers,  how  do 
we  find  their  sum?         ■lm^i^ii&  uemi  •!^'  bi4  i^«  «£.  sso^pwifi  sviii  nt:  ■ 


Art.   1.  ^ECVPITULATION.  53 

field,  tlieir  stun  would  express  the  number.  Also,  the  sum  of  2 
iinifs  and  6  iem,  is  neither  8  units  nor  8  tens  ;  but  if  we  reduce  the 
6  tens  to  units,  their  sum  would  be  68  units.  Thus,  we  perceive 
that  2ve  cannot  add  several  numbers  together,  unless  they  can  be  classi- 
jied  under  one  order  in  which  one  name  is  common  to  them  all. 

Obs.  8.  When  we  have  giv^en  several  numbers,  to  find  their 
sum — Add  them  together. 

Ex.  1.  There  are  two  numbers  ;  one  is  144,  and  the  other  is  62. 
What  is  their  sum?  Ans.  206. 

2.  A.  has  1200  dollars,  and  B.  has  430  dollars.  How  many  dol- 
lars have  both?  Ans.   1680  dollars. 

3.  The  distance  A.  has  traveled  is  180  miles;  B.  has  traveled 
114  miles.     How  many  miles  have  both  traveled?         Ans.  294. 

Obs.  9.  It  may  be  asked  by  some — Why,  in  Addition,  we  car- 
ry 1  for  every  ten,  instead  of  one  for  every  eight,  nine,  twelve,  or 
some  other  number?  The  answer  is — because  our  systefn  of  notation 
increases  in  a  ten-fold  ratio  ;  consequently,  as  often  as  ive  have  ten  of 
one  order,  we  must  have  1  of  the  next  higher  order.  Had  the  base  of 
our  system  of  notation  been  eight,  7une,  twelve,  or  any  other  number, 
instead  of  ten,  we  should  have  carried  1  for  every  eight,  nine,  twelve^ 
or  whatever  other  number  was  the  base  of  our  scale  of  notation. 
The  same  remark  applies  to  Subtraction,  Multiplication,  and  Divi- 
sion. 

Obs.  10.  Subtraction. — The  numbers  must  all  be  of  the  same 
name,  or  hind  in  Subtraction  as  ivell  as  in  Addition.  Thus:  if 
I  have  in  a  field  13  cherry  trees,  and  5  pear  trees,  my  taking 
the  pear  trees  away  does  not  diminish  the  number  of  cherry 
trees,  although  it  lessens  the  number  of  trees  in  the  lot. 

Again  :  we  cannot  take  2  tmits  from  8  fens,  unless  we  first  reduce 
the  tens  to  units.  Thus,  we  perceive  that  we  cannot  subtract  two 
^lumbers,  unless  they  have  a  name  common  to  both: 

It  is  not  absolutely  necessary,  in  Subtraction,  that  we  place  the 
greater  number  above  the  other.     This  is  done  only  for  convenience. 

Obs.  11.  When  we  have  given  the  sum  of  two  numbers,  and 
either  number,  to  find  the  other — From  the  sum  s?ibtract  the  given 
number. 

Demonstrate  this  rule.  Why,  in  Addition,  do  we  carry  1  for  every 
ten,  instead  of  1  for  every  eight,  twelve,  or  some  other  number?  Had 
the  lase  of  our  Byslein  of  notiitioa  been  eight,  twelve,  or  some  other 
nuMiber  iustead  of  ten,  !io\v  often  would  we.  have  carried  1  ?  Explain 
why,  ill  Rublraction.  the  numbers  ninst  all  be  of  the  sanje  name,  or  kind. 
What  is  necessary  in  order  to  subtract  two  numbers?  Is  it  necessa- 
ry in  Subtraction  that  we  place  the  greater  number  above  the  other?  Why 
then  is  it  done?  W  hen  we  have  given  the  sum  of  two  numbers,  and  either 
number,  how  do  we  find  the  other  number?     Demonstrate  this  rule. 


54  COMMON    ARITHMETIC.  Scct.  VI. 

4.  The  sum  of  two  numbers  is  40,  and  one  of  ihem  is  18.  What 
is  the  other  number?  Ans.  22. 

5.  Two  men  have  1000  dollars  ;  A.  lias  572  dollars.  How  many 
dollars  has  B.?  Ans.  428. 

Obs.  12.  When  the  difference  and  greater  of  two  numbers  are 
given  to  find  the  lesser  number — From  the greaternumber  subtract  the 
difference, 

6.  The  greater  of  two  numbers  is  53,  and  their  difference  is  14. 
Required — the  lesser  number.  Ans.  39. 

7.  A.  has  1000  dollars,  which  is  346  dollars  more  than  B.  has. 
How  many  dollars  has  B.?  Ans.  654. 

8.  A.  has  traveled  300  miles,  which  is  163  miles  farther  than  B. 
has  traveled.     How  far  has  B.  traveled?  Ans.   137  miles. 

Obs.  13.  Given,  the  dfference  w[i^l€sser  of  two  numbers,  to  find 
ihe  greater — Add  the  dfference  and  lesser  number  together, 

NqfE. — This  is  the  same,  in  reality,  as  the  proof  of  Subtraction. 

9.  The  difference  of  two  numbers  is  7,  and  the  lesser  number  is 
13.     What  is  the  greater  number?  Ans.  20. 

1  0.  A.  has  560  dollars,  and  B.  has  40  dollars  more.  How  much 
has  B.?  Ans.  600  dollars. 

11.  A.  went  140  miles,  and  B.  went  60  miles  farther.  How  far 
did  B.  go?  Ans.  200  miles. 

Obs.  1 4.  Given  the  sum  and  difference  of  two  numbers,  to  find 
those  numbers — From  their  sum  subtract  their  difference,  and  half  the 
remainder  ivill  be  the  lesser  number. 

12.  The  sum  of  two  numbers  is  16,  and  their  difference  is  4.  Re- 
quired— the  numbers.  Ans.  6  and  10. 

13.  The  sum  of  the  ages  of  two  men  is  66  years,  and  one  of  them 
is  20  years  older  than  the  othel*.     Required — their  ages. 

Ans,  43  and  23  years. 

14.  Two  men  together  have  500  dollars,  and  one  of  them  has  240 
dollars  more  than  the  other.     How  many  dollars  have  both? 

Ans.  One  has  370  dollars,  and  the  other  130  dollars. 

Obs.  15.  Multiplication.— 'As  Multiplication  is  the  repeated  ad- 
dition of  the  same  number  to  it^iclf,  it  may  be  thought  by  some  that 
the  multiplier  can  sometimes  bo  a  concrete  number,  (as  we  cannot 
add  numbers  of  different  names  together).     But  the  learner  must 


When  the  difference  and  greater  of  two  nunibersare  given,  how  do  we  find  the 
lesser  number?  Demonstrate  this  rule.  When  we  have  given  the  difference  and 
lesser  of  two  numbers,  how  do  we  find  the  greater?  Demonstrate  this  rule 
When  we  have  given  the  sum  and  difference  of  two  numbers,  how  do  we  find 
the  numbers?  Demonstrate'  this  rule.  Explain  why  the  multiplier  is  always 
an  abstf;i(;t  nun\ber.     Wlien  vye  have  given  the  product  of  two  factors,  and 


Art.    1.  RECAPITULATION.  55 

recollect  that  the  multiplicand  is  the  number  to  be  added,  and  that 
the  multiplier  only  expresses  how  many  times  it  is  to  be  added,  or  re- 
peated. Hence,  the  multiplier  is  always  an  abstract  number.  To 
multiply  acres  by  dollars,  or  cents  by  yards,  &c.,  is  sheer  nonsense. 
For  th  method  of  analyzing  concrete  questions,  see  Sect.  IV,  Art. 
2,  Ex.  1. 

Obs.  \Q.  Given  the  p-oduct  of  two  factors,  and  either  factor,  to 
find  the  other  factor — Divide  the  prodicct  hy  the  given  factor.  (Sect. 
v.,  Art.  2,  Obs.  10,) 

15.  The  product  of  two  factors  is  48,  and  one  of  them  is  6. 
What  is  the  other  factor?  Ans.  8. 

16.  The  product  of  two  factors  is  576,  and  one  of  them  is  48. 
What  is  the  other  factor?  Ans.  12. 

To  prove  Multiplication  by  casting  out  the  9's — 

Obs.  17.  Cast  the  9's  out  the  multiplier  and  imdtiplicand ;  mul- 
tiply their  excesses  together,  and  cast  the  9'^  oui  of  their  product ; 
then,  if  the  excess  of  9'5  in  this,  is  the  same  as  the  excess  of  9'5  in 
the  total  product,  or  answer  to  the  question,  the  work  is   correct. 

This  method  of  proof  depends  upon  A  property  of  the  number  9, 
viz: 

Any  number  divided  by  9  will  leave  the  same  remainder  as  the 
sum  of  its  digits,  or  figures,  divided  hy  9. 

JJemoTistration. — Take  anv  number,  as  876.  This  separated  into 
its  nume  ical  parts,  equals  800  -|-  70  +  6.  But  800  =  8  X  100  =  8 
X  (99  +  1)  =  8  X  99  +  8.  Also,  70  =  7  X  9  -}-  7.  Hence,  876 
=  8  X99-)-8  +  (7X9  +  7)  +  6=:=8  X  99  +  (7  X  9)  +  8  +  7 
+  6,  and  876  ~-  9  =  (8  X  99  +  7  X  9  +  8  -f-  7  +  6)  ^  9.  But 
8  X  99  +  (7  X  9)  is  evidently  divided  by  9  without  a  remainder, 
(because  9  is  one  factor  of  this  expression) ;  therefore,  876-^9  wiU 
leave  the  same  remainder  as  (8-}-  7  -j-  6)  -f-  9.  The  same  method 
of  reasoning  will  apply  to  any  other  number. 

Now,  from  this  demonstration,  the  reason  of  the  rule  is  evident. 
Because,  if  we  reject  the  9's  from  any  number,  and  also  reject  the 
9's  from  the  several  parts  of  the  same  number,  add  the  latter  ex- 
cesses together,  and  reject  the  9's  from  their  sum,  this  latter  excess 
ought  to  be  equal  to  the  excess  of  9's  in  the  number  itself;  the 
whole  being  equal  to  the  sum  of  all  its  parts. 

Now,  in  Multiplication,  the  product  is  a  number,  of  which  the 
multiplicand  is  a  part,  taken  as  many  times  as  there  are  units  in  the 

eilher  facter,  how  do  we  find  the  other  factor?  Demonstrate  this  rule.  How 
do  we  prove  multiplication  by  casting  out  the  9's?  Upon  what  dops  this  meth- 
od of  proof  depend?  What  is  this  property?  Demonstrate  it.  Show  from  the 
demonstration  why  the  rule  is  evident.    How  is  this  applied  to  Multiplication? 


8(>  COMMON    ARITHMETIC.  ScCt.  YI. 

multiplier.  Hence,  if  we  multiply  the  multiplier  by  the  excess  of 
9*s  in  the  multiplicand,  the  excess  of  9's  in  the  product  ought  to  be 
equal  to  the  excess  of  9's  in  the  total  product,  or  answer  to  the 
question.  But  it  will  produce  the  same  result  to  multiply  the  ex- 
cess of  9's  in  the  multiplicand  by  the  excess  of  9's  in  the  multi- 
plier, and  reject  the  9's  from  this  product.  Hence,  the  rule  is 
correct. 

To  cast  out  the  9's  in  any  number,  we  add  together  the  digits  of 
that  number,  and  as  often  as  we  obtain  9,  reject  it,  take  the  remain- 
der and  proceed  as  before.  Thus,  to  cast  the  9's  out  of  1657324, 
we  say  4  and  2  are  6,  and  3  are  9;  (rejecting  the  9,)  7  and  5  are  12  ; 
(rejecting  the  9,)  3  and  6  are  9  ;  (rejecting  the  9,)  we  find  the  ex- 
cess in  the  whole  number  to  be  1 . 

Note. — This  property  of  the  number  9,  belongs  to  no  other  digit  except  3. 

17.  Multiply  278  by  745,  and  prove  the  operation. 
Operation. 

278  Excess  of  9*s  in  the  multiplicand  is  8. 

745  "         "         *'      multiplier  is  7. 

*'         *]         "      product  of  8X7  is  2. 

1390  *•         "         "      whole  product  is  2. 

1112  Hen'f,  the  operation  is  correct. 
1946 


207110 
It  is  customary  to  write  the  excesses  in  the  four  spaces  of  a  cross; 
the  excesses  of  the  two  factors  being  placed  above  and       \  8  / 
below,  and  the  other  excesses  at  the  right  and  left,  thus  :      ^y\2 
Then  if  the  excesses  at  the  right  and  left  are  alike,  the      /  7\ 
work  is  correct. 

18.  What  is  the  product  of  4682  by  378?  Ans.   1769796. 

19.  What  is  the  product  of  4781  by  6213?      Ans.  29704353. 

20.  What  is  the  product  of  37682  by  26571? 

Ans.   1001248422. 
Obs.  18.     When  we  have  given  the  cost  of  1,  to  find  the  cost  of 
a  given  quantity,  either  more  or  less — Multiply  the  cost  and  quantity 
together. 

Remark. — The  term  quantity  applies  to  any  thing  capable  of  increase  or 
diminution — as  numbers,  lines,  cloth,  &-c. 

Note. — The  learner  must  not  infer  from  the  above  that  we  multiply  two 
concrete  numbers  together,  because  we  say,  multiply  together  the  cost  and 


How  do  we  cast  the  9's  out  of  a  number?  Does  this  property  of  the  number 
U  belong  to  any  other  digit?  When  we  have  given  the  cost  of  1,  how  do 
we  find  the  cost  of  a  given  quantity,  either  more  or  less?  To  what  does  the 
term  quantity  apply  ? 


Art.  1,  RECAPITULATION*  57 

quantity.  We  express  the  rule  in  this  way  for  the  sake  of  brevity,  but  the 
multiplier  should  always  be  abstract.  (Obs.  15,  and  Sect.  IV,  Art.  2,  Obs. 
6,Rem.  1.) 

21.  If  1  busliel  of  apples  cost  42  cents,  how  much  will  25  bushels 
cost?  Ans.   1050  cents. 

22.  If  1  stove  costs  25  dollars,  how  much  will  19  stoves  cost? 

Ans.  475  dollars. 

23.  If  1  book  costs  84  cents,  how  much  will  216  books  cost? 

Ans.   18144  cents. 

Obs,  19.  If  we  add  a  unit  to  the  multiplier,  it  will  produce  the 
same  result  as  to  add  the  multiplicand  to  the  product ;  if  we  add 
two  units,  the  result  will  be  the  same  as  to  add  the  multiplicand 
iwice  to  the  product ;  and  universally,  to  add  any  number  to  the  mul- 
tipller,  iTKreases  the  'produd  as  many  times  the  multiplicand  as  there 
are  units  in  the  number  added. 

Again,  if  we  subtract  a  unit  from  the  multiplier,  it  will  produce 
the  same  result  as  to  subtract  the  multiplicand  from  the  product;  to 
subtract  two  units,  th^  result  will  be  the  same  as  to  subtract  twice  the 
multipUcand  from  the  product;  and  universally,  to  subtract  any  num- 
ber from  themidtiplier,  diminishes  the  product  as  many  times  the  mul- 
tiplicand, as  there  are  units  in  the  number  subtracted. 

It  is  for  these  reasons  that  the  proof  in  Multiplication  is  correct. 
(Sect.  IV.,  Art.  2,  Obs.  7.) 

Remark. — The  learner  will  bear  in  mind  that  in  all  cases  where  the 
multiplier  is  unity,  or  1,  the  product  is  equal  to  the  multiplicand  ;  when 
the  multiplier  is  greater  than  1,  the  product  is  greater  than  the  multipli- 
cand ;  and  when  tlie  multiplier  is  less  than  1,  the  product  is  less  than  the 
multiplicand. 

Obs.  20.  Division. — Division  has  been  defined  by  many  au- 
thors of  Arithmetic  as  a  short  method  of  performing  many  subtractions  y 
when  the  numbers  to  be  subtracted  are  all  equal  ;  and  as  a  consequence 
it  is  urged  that  the  divisor  is  always  of  the  same  name  as  the  divi- 
dend, (because  we  cannot  subtract  numbers  of  different  names  from 
each  other,)  and  that  the  quotient  is  always  an  abstract  num- 
ber, (because  it  expresses  the  number  of  times  the  subtractions  have 
been  performed.) 

This,  however,  is  not  always  the  case,  because  we  very  often 
have  two  concrete  numbers  given — one  as  a  dividend,  and  the  other 

If  we  add  a  unit  to  the  multiplier,  what  effect  does  it  have  on 
the  product?  If  we  add  two  units  what  eifect  does  it  have?  What 
inference  is  deduced  from  this?  What  effect  does  it  have  on  the  product  to 
subtract  a  unit  from  the  multiplier?  Two  units?  What  inference  is  deduced 
from  this?  If  the  multiplier  is  1,  to  whnt  is  the  product  equal?  If  the  mu!- 
tipiier  is  irrei-iter  th.au  1,  to  what  is  tiic  i>roduct  equal?  If  th"  ma:'.i:>';i'.^f  >s 
less  than  i,  to  what  is  the  product  equal?  How  is  oivic.iou  iiuiiuou  by;i.a- 
ny  authors?    Wiiat  inference  i«deduceti  from  this?    Is  this  always  this  case? 


58  COMMON    ARITHMETIC.  Sect.  VI, 

as  a  divisor,  which  are  of  entirely  different  names.  For  instance, 
suppose  it  were  required  to  divide  100  dollars  equally  between  4 
men,  and  tell  how  many  dollars  each  would  receive.  Now,  how 
are  vre  to  take  4  men  from  100  dollars ,  and  how  many  dollars  will 
remain  after  the  operation  has  been  performed  ?  Why,  the  very 
idea  of  such  a  thing  is  absurd  and  preposterous  in  the  extreme. 

The  learner  will  bear  in  mind  that  the  dividend  is  equal  to  the 
product  of  the  divisor  and  quotient,  plus  the  remainder,  if  any. — 
(Sect.  Y.,  Art.  2,  Obs.  9.)  Hence,  the  divisor  or  quotient  must 
one  of  ihem  always  be  of  the  same  name  as  the  dividend,  since 
the  product  must  always  be  of  the  same  name  as  the  multiplicand. 
(Sect  IV.,  Art  2,  Obs.  6.) 

Kow  in  the  above  example  the  dividend  is  dollars,  and  the  an- 
swer required  is  dollars  ;  and  we  reason — if  4  men  receive  100  dol- 
lars, one  man  ivill  receive  one-fourth  o/*  100  dollars.  But  one-fourth 
of  one  hundred  dollars  is  25  dollars,  and  it  cannot  be  any  thing 
else.  Here,  our  divisor  is  abstract,  whilst  our  dividend  and 
quotient  are  both  concrete.  This  example  could  not  be  performed 
by  Subtraction;  hence,  the  above  definition  of  Division  cannot  be 
true  in  all  cases.  The  divisor  is  abstract,  and  the  dividend  and 
quotient  are  concrete;  hence,  the  inferences  deduced  from  this  defi- 
nition are  incorrect. 

Obs.  21.  Given  the  divisor  and  quotient  to  find  the  dividend — 
Multiply  the  divisor  and  quotient  together,  and  to  the  product  add  the 
remainder,  if  amj. 

24.  The  divisor  of  a  certam  number  is  15,  and  the  quotient  is  6. 
Kequired — the  number.  Ans.  90. 

25.  If  the  divisor  is  24,  and  the  quotient  36,  what  is  the  divi- 
dend? Ans.  864. 

26.  If  the  divisor  is  48,  the  quotient  15,  and  the  remainder  37, 
what  is  the  dividend?  Ans.  757. 

Obs.  22.  Griven  the  dividend  and  quotient  to  find  the  divisor — 
Subtract  the  remainder  [if  any ^)  from  the  dividend,  and  divide  the  re- 
sidt  by  the  quotient. 

27.  If  the  dividend  is  456,  and  the  quotient  is  8,  what  is  the  di- 
visor? Ans.  57. 

28.  If  the  dividend  is  108,  and  the  quotient  9,  what  is  the  divi- 
sor? Ans.   12. 

29.  If  the  dividend  is  1738,  the  quotient  144,  and  the  remainder 
8,  what  is  the  divisor?  Ans.   12. 

Why  not?  Give  an  example.  To  what  is  the  dividend  equal?  What  in- 
ference is  deduced  from  this?  Why?  In  the  example  given  what  is  the  div- 
idend? What  answer  is  required?  How  do  we  reason?  Of  virhat  name  is 
our  divisor  in  tliis  case?  Our  dividend  and  quotient?  CoMJd  this  ex^nn- 
T)le  have  been  performed  by  Subtraction?  What  do  we  couclude  from  this? 
When  we  have  given  the  divisor  and  quotient,  how  do   we  find  the  dividend? 


Art.   1 .  RKCAHTULATION.  59 

30.-  If  the  dividend  is  4829,  the  quotient  37,  and  the  remainder 
19,  what  is  the  divisor?  Ans.   130. 

Obs.  23.  To  prove  division  by  casting  out  the  9's — Cast 
the  9's  Old  of  ike  divisor  and  quotient ;  multiply  their  excesses  together ^ 
and  cast  the  9'6'  out  of  their  product ;  cast  the  d's  out  of  the  remain' 
der,  add  this  excess  to  the  last,  and  cast  the  9'5  out  of  their  sum.  ; 
and  if  this  latter  excess  is  equal  to  the  excess  of  9's  in  the  dividend,  ike 
work  is  correct. 

This  Rule  is  demonstrated  in  the  same  manner  as  the  rule  for 
proving  Multiplication  by  casting  out  the  9's,  the  dividend  being  the 
number  separated,  and  the  parts  being  the  divisor,  (or  quotient,^ 
taken  as  many  times  as  there  are  units  in  the  quotient,  (or  divisor,) 
and  remainder. 

It  may  be  proper  to  remark  that  Addition  and  Subtraction  can  al- 
so be  proved  by  casting  out  the  9's,  but  as  the  method  of  proof  giv- 
en in  Sects.  II.  and  III.  is  full  as  short,  and  easier  to  be  understood, 
we  shall  let  them  suffice  at  present.  An  ingenious  scholar,  howev- 
er, can  study  out  the  proof  from  the  demonstration  and  remarks  un- 
der Obs.  17,  if  he  thorougly  understands  the  subject  thus  far. 

31.  Divide  207748  byV45,  and  prove  the  operation. 
Operation. 

745)207748(278      The  excess  of  9's  in  the  divisor  is  7 

1490  *'         **         "         "     quotient  is  8 


6874 
5215 

n 

it 
,t 

t( 
a 

tt 

n 

t( 
tt 

"  product  of  8X7  is  2 
"     remainder  is           8 

6598 
5960 

"  sum  of  8  -f-  2  is  1 
"     dividend  is              1 

638  Hence,  the  work  is  correct. 

32.  Divide  456789  by  365.  Ans.  1251,  and  174  rem. 

33.  Divide  764218  by  213.  Ans.  3587,   andl87  rem. 

34.  Divide  932684  by  416.  Ans.2242,andl2  rem. 

35.  Divide  897653  by  321.  Ans.  2796,  and  137  rem. 
Obs.  24.     When  we  have  given  a  quantity  either  more  or  less,  to 

find  the  cost  of  unity,  or  1 — Divide  ike  cost  by  the  quantity. 

36.  If  8  pounds  of  coffee  cost  100  cents,  how  much  is  that  per 
pound?  Ans.    12^-  cents. 

Demoustrate  this  rule.  When  we  have  given  the  dividend  and  quotient,  how 
do  we  find  the  divisor?  Demonstrate  this  rule,  dew  do  we  prove  divis- 
ion by  casting  out  the  9's?  How  is  this  rule  demonstrated?  Which  is  the 
number  separated?  Which  are  tlie  parts?  Can  Addition  and  Subtraction  be 
proved  by  casting  out  the  9's?  Can  you  study  out  the  rule?  When  we  have 
given  a  quantity  either  mora  or  less,  and  its  cost,  how  do  we  find  the  cost  of  1?^ 


60  COMMON   ARITHMETIC.  Scct.   VI. 

37.  If  24  stoves  cost  576  dollars,  how  much  is  that  apiece? 

Ans.  24  dollars. 

38.  If  124  horses  cost  8308  dollars,  how  much  is  that  apiece? 

Ans.  67  dollars. 

a.  When  we  have  given  the  cost  of  a  quantity,  and  the  cost  of 
unity,  or  1,  to  find  the  quantity — Divide  the  cost  of  the  quantity  hythe 
cost  of  unOyi  or  1. 

39.  How  many  acres  of  land  can  I  buy  for  3834  dollars,  at  27 
dollars  per  acre?  Ans.  142. 

40.  How  many  stoves  can  I  buy  for  336  dollars,  at  28  dollars 
apiece?  Ans.  12. 

Obs.  25.  Although  Subtraction  and  Division  both  separate  num- 
bers into  ^ar^5,  the  learner  will  observe  that  there  is  a  great  differ- 
ence between  them.  In  Division  the  parts  are  always  equal,  (being 
factors  of  the  dividend;)  but  in  Subtraction,  the  parts  may  be  either 
equal  or  unequal.  Thus  24  may  be  separated  into  4  parts  by  Divi- 
sion, each  of  which  is  6  ;  and  likewise  it  may  be  separated  into  4 
parts  by  Subtraction,  the  parts  being  7,  3,  9,  5,  or  8,  6,  4,  6,  or 
9,  8,  5,  3,  &c.  But  the  former  parts  (factors,)  we  multiply  together 
to  produce  24,  (6  X  4  =  24,)  whilst  we  add  the  latter  to  obtam  the 
same  result.  (7  +  3  +  9  -f  5  =  24,  &c.)  (Sect.  IV.,  Art.  4,  Obs. 
4,  Rem.  2.) 

Obs.  26.  It  is  plain  from  the  nature  of  Division,  that  the  value  of 
the  quotient  depends  both  on  the  divisor  and  dividend.  Because,  it 
is  self-evident,  that  with  the  same  dividend,  the  greater  the  divisor, 
the  less  will  he  tJie  quotient,  and  the  smaller  the  divisor,  the  greater 
will  be  the  quotieTit.     Hence — 

a.  If  we  increase  our  divisor  we  diminish  our  quoile7it ;  and  con- 
versely,  if  we  diminish  our  divisor,  ive  increase  our  quotient,  if  the 
dividend  remains  unaltered.     Because — 

h.  If  our  divisor  is  unity,  or  1,  the  quotient  is  equal  to  the 
dividend  ;  if  our  divisor  is  greater  than  1 ,  the  quotient  is  less  than 
the  dividend;  and  if  our  divisor  is  less  than  1 ,  the  quotient  is  greater  thna 
the  dividend. 

€.  It  is  also  self-evident,  that  if  we  increase  the  dividend,  we 
increase  the  quotient ;  and  if  we  diminish  the  dividend,  we  diminish 
the  quotient,  if  the  divisor  remains  unaltered.     Because — 

When  we  have  given  the  cost  of  a  quantity,  and  the  cost  of  unity,  how  do  we 
find  the  quantity?  Explain  the  difference  between  Subtraction  and  Division. 
Give  an  example.  Upon  what  does  the  value  of  the  quotient  depend?  Why? 
What  inference  is  deduced  from  this  fact?  If  our  divisor  is  unity,  or  1,  to  what 
is  the  quotient  equal?  If  our  divisor  is  greater  than  unity,  wh?it  is  the  value 
of  the  quotient?  If  our  divisor  is  less  than  unity,  what  is  the  value  of  iljc 
quotient?  What  effect  does  it  have  upon  the  quotient  to  increase  or  decrease 
our  dividend,  without  altering  the  divisor? 


Art.    1.  RECAPITULATION.  Ql  ^ 

d.  When  the  dividend  and  divisor  are  equals  the  quotient  is  unity, 
or  1  /  when  the  dividend  is  greater  than  the  divisor,  the  quotient  is 
greater  than  1  /  and  when  the  dividend  is  less  than  the  divisor,  the  quo- 
tient is  less  than  1. 

Obs.  27.  From  these  remarks,  it  is  evident  that  it  produces  the 
same  effect  en  the  quotient,  to  multiply  the  divisor  by  any  number, 
as  to  divide  the  dividend  by  the  same  number  ;  and  also,  it  produces 
the  same  effect  on  the  quotient  to  divide  the  divisor  by  any  number, 
as  to  multiply  the  dividend  by  the  same  number.  From  these  facts, 
we  deduce  the  following  considerations  : 

a.  1st.  To  divide  the  divisor  by  any  number,  or  to  multiply  the 
dividend  by  the  same  number,  is  in  effect  multiplying  the  quotient  by  this 
number.     Thus: 

4  is  contained  in  12  3  times. 

4  —  2         *'  "12  6  times,  or  3  X  2  times. 

4         **  *'     12  X  2         6  times,  or  3  X  2  times. 

h.  2d.  To  multiply  the  divisor  by  any  number,  or  to  divide  the  di- 
vidend by  the  same  number,  is  in  effect  dividing  the  quotient  by  this 
number.     Thus: 

3  is  contained  in  12  4  times. 

3X2         "  "12  2  times,  or  4  -^  2  times. 

3         "  *'     12-f-2         2  times,  or  4 -^  2  times. 

Obs.  28.   To  multiply  or  divide  both  the  divisor  and  dividend  by 
the  same  number,  does  not  alter  the  quotient.     Thus: 
8  is  contained  in  16  2  times. 

8X2         "  '*     16  X  2         2  times. 

8-~2         "  "     16-r-2         2  times. 

Obs.  29.     If  we  add  the  same  number  to  both  the  divisor  and  divi- 
dend, ice  diminish  the  quotient ;  and  if  we  subtract  the  same  number 
from  both  the  divisor  and  dividend,  we  increase  the  quoiietit.     Thus  : 
6-r-2=  3.  Ana  12-^4=  3. 

64-2—2  +  2  =  2.  12—  2-^(4— 2)=5. 

Remark. — In  each  of  these  cases  the  divisor  is  supposed  to  be  less  than  the 
dividend.  When  the  divisor  and  dividend  are  equal  it  does  not  alter  the  quo- 
Why  is  this  correct?  What  fact  is  evident  from  these  remarks?  What  is 
the  first  consideration  we  notice  from  this  fact?  Give  an  example.  Wh?it  is 
the  second  consideration?  Give  an  example.  What  effect  does  it  have  upon 
the  quotient  to  multiply  or  divide  both  the  divisor  and  dividend  by  the  same 
number?  Give  an  example  If  we  add  the  same  number  to  both  the  divisor 
and  dividend,  what  effect  does  it  hav^i  upon  the  quotient?  If  we  subtract  the 
same  number  from  both  the  divisor  and  dividend,  what  efTect  does  it  have  on 
the  quotient?  Give  examples.  In  the  last  two  cases,  what  i^  the  vaiuH  of  the 
divisor,  compared  with  that  of  the  dividend?  If  tha  divisor  and  dividend  are 
equal,  what  effect  does  it  have  upon  the  quotient  to  add  the  sume  number  to, 
or  subtract  the  same  number  from  both? 


62,  COMMON   ARITHMETIC.  Sect.   VI. 

tient  to   add  the  same  number  to,  or  subtract  it  from  both.     Also When 

the  divisor  is  greater  than  the  dividend,  the  converse  of  the  above  propositions 
is  true. 

Obs.  30.     If  a  given  number  he  both  midtipUed  and  divided  hy  the 

same  number  >  the  final  result  idll  be  the  original  number.    Thus 

8  X  7  =  56.  66  -^  7  =  8. 

Article  2.     Contractions  and  Abbreviations. 

Note. — In  addition  to  the  rules  for  contracting  the  operations  in  Multi- 
plication and  Division,  given  in  Sects.  IV.  and  V.  the  following  may  also  be 
of  advantage  in  some  particular  cases. 

Obs.  1.  As  to  multiply  by  10,  100,  1000,  (fee,  we  have  merely  to 
annex  ciphers,  to  the  multiplicand,  [Sect.  IV.,  Art.  4,  Obs.  1.]  it 
follows,  that  if  a  multiplier  is  an  exact  part  of  100,  1000,  &c.,  we 
can  annex  two,  three,  or  more  ciphers,  if  necessary,  to  the  multi- 
plicand, and  then  divide  it  by  such  a  number  ;:s  the  multiplier  is  a 
part  of,    100,  1000,  &c.     Thus  : 

To  multiply  by  25 — Anmx  two  ciphers,  and  divide  by  4  :  because 
25  is  one-fourth  of  100. 

To  multiply  by  50— Annex  two  cijyhers,  and  divide  by  2:  because 
60  is  one-half  of  100. 

To  multiply  by  125 — Annex  three  ciphers,  and  divide  by  8:  be- 
cause  125  is  one-eio-hth  of  1000. 

Ex.  1.  Multiply  48  by  25.  Ans.   1200. 

J.  Multiply  72  by  25.  Ans.   1800. 

3.  Multiply  649  by  25.  Ans.   16225. 

4.  Multiply  288  by  50.  Ans.   14400. 

5.  Multiply  897  by  50.  Ans.  44850. 

6.  Multiply  462  by  125.  Ans.  57750. 

7.  Multiply  3426  by  125.  Ans.  428250. 

8.  Multiply  7894  by  125.  Ans.  986750. 

9.  Multiply  478264  by  12348. 

Operation. 
478264 
12348 


1434792 
5739168 
22956672 

Ans.  6905603872. 


If  the  divisor  is  greater  than  the  dividend,  what  is  the  effect?  If  a  number 
be  multiplied,  and  the  product  divided  by  the  same  number,  what  effect  does  it 
liavo  upon  the  final  result?  Give  an  example.  How  do  we  multiply  by  10, 
100,  lOO'J,  &c.?  What  inference  is  deduced  from  this?  How  can  we  multi- 
ply by  25?    Why  is  this  correct?     By  50?     Why  is  this  correct?     By  123? 


Art.    2.  CONTRACTIONS    AND    ABBREVl^TlOXg.  63 

"We  first  multiply  by  3,  and  then  multiply  this  product  by  4,  be- 
cause 3X4=12,  and  consequently  4  times  3  times  any  number  is 
evidently  twelve  times  that  number.  Kext,  we  multiply  this  latter 
product  by  4,  because  12  X  4  =  48,  and  place  the  first  figure  of  the 
product  under  the  8,  because,  in  reality,  we  multiply  the  multipli- 
cand by  4^.  The  first  figure  of  our  second  partial  product  we  place 
under  the  2,  because,  in  reality,  we  multiply  the  multiplicand  by 
12.  'J'hc  several  results  are  added  together,  as  usual.  The  ope- 
ration is  proved  by  casting  out  the  9's. 

10.  Multiply  14246  by  819.  (81  =  9X9.)      Ans.   11667474. 

11.  Multiply  327436  by  126721.  (21  =7  X3;   126  =  21  X  6.) 

Ans.  41493017356. 

12.  Multiply  9476245  by  648963.  (63  =  9  X  7  ;  64  =  8  X  8.) 

Ans.  6149732383935. 
Obs.  2.     The  learner  will  perceive  from  these  examples,  that  ii 
makes  no  difference  which  figure  of  the  multiplier  we  multiply  hy  first, 
provided  we  place  the  first  figure  of  the  product  directly  under  the  fig- 
ure by  ichich  we  multiply. 

13.  Multiply  326  by  241. 

Operation. 

By  this  method  we  number  the  figures  of  3^ 2^ 6^ 

each  factor,  commencing  (at  the  right  hand,)  2^4^° 

with  1  in  the  multiplicand,  and  0  in  the  mul- 

tipher.     This   is   done  to  assist   the  memory.   Ans.  7^8'' 5^ 6^ 6*. 
These  small  figures  we  call  Exponents. 

We  number  each  figure  of  the  product,  (our  first  exponent  being 
1,)  ^0  shoio  which  figures  we  multiply  together.  Thus,  we  wish  in  the 
first  place  to  find  a  figure  in  the  product,  the  exponent  of  which  is 
1.  "ifow  we  multiply  those  figures  together,  the  sum  of  whose  ex- 
ponents is  equal  to  the  exponent  required  in  the  product :  the  expo- 
nent of  1  is  0,  and  the  exponent  of  6  is  1 ;  0  -f-  1  =  1 ;  then  we  mul- 
tiply 6  by  1,  and  6  X  1  =  6,  the  first  figure  of  the  product.  Next 
we  wish  to  find  a  figure  of  the  product  whose  exponent  is  2.  The 
exponent  of  1  is  0,  and  that  of  2  is  2;  0  -j-  2  =  2:  also  the  exponent 
of  4  is  1,  and  that  of  6  is  1;  1  -[-  1  ==  2;  then  2  X  1  =  2,  and  6  X 
4  =  24;  24+2  =  26.  We  set  down  the  6  and  carry  the  2  as 
usual.  Now  we  wish  a  figure  in  the  product  whose  exponent  is  3: 
0  -f  1  =  3;  1  +  2  =  3;  and  2  +  1  =  3;  then  3  X  1  -f-(2  X  4)  + 
(6X2)  =  23,  and  2  to  carry  makes  25.     Then  5  is  our  third  figure 


Why  is  this  correct?  Explain  tiie  operation  of  Kx.  9.  Whiit  conclusioa  is 
drawn  from  this  and  the  three  lollcvviDg  exanip!c:i?  What  is  our  first  step  in 
the  operation  of  Ex.  13?  Why  is  this  doiie?  What  are  these  sm<ill  ficrnres 
called?  What  is  onr  fi"?t  exponent  in  tlie  pro:ii.ict?  Why  do  we  nuMjb^r  the 
product?  What  do  we  wish  to  obtain  lirst?  Which  figures  do  we  multiply  to- 
gether? 

'  '4 


64  COMMON    ARITHMETIC.  Scct.   VI. 

in  the  product,  and  2  to  carry.  The  next  figure  in  the  product  must 
have  4  for  its  exponent:  1  -j-  3  =  4,  and  2  -|~  2  =  4;  then  3X4  + 
(2X2)  =  16,  and  2  to  carry  makes  18;  we  set  down  8  and  carry  1. 
Our  next  exponent  in  the  product  is  5;  2  -)-  3  =  5;  then  3X2=6, 
and  1  to  carry  makes  7  as  our  next  figure  in  the  product,  and  as 
the  sum  of  no  two  exponents  exceeds  5,  we  conclude  that  our  work 
is  done. 

Obs.  3.  By  examining  this  operation  attentively  we  notice  the 
following  considerations: 

1st.  We  add  but  two  exponents  together  at  a  time,  one  of  which  be- 
longs to  each  factor. 

2d.  We  multiply  those  figures  together,  the  sum  of  whose  exponents 
is  equal  to  the  exponent  sought,  add  their  several p)roducts  together  men- 
tally^ and  set  down  and  carry  as  usual. 

Obs.  4.  The  only  difference  between  this  and  the  usual  method 
of  multiplying  is,  that  by  this  method  the  operation  is  performed 
mentally,  whilst  by  the  other  it  is  all  written  upon  the  slate  ;  the 
mental  exercise  and  discipline  of  mind,  (no  contemptible  considera- 
tions,) being  the  chief  advantages  gained  by  this  manner  of  operating. 

14.  Multiply  379  by  647.  Ans.  245213. 

15.  Multiply  1846  by  1234.  Ans.  2277964. 

16.  Multiply  27345  by  97216.  Ans.  2658371520. 

17.  Multiply  894832  by  687219.  Ans,  614945552208. 

Remark.— We  liave  already  learned  ho,v  to  multiply  by  25..  50  and  125,  and 
it  is  evident  that  we  can  divide  by  these  numbers,  by  merely  reverijiug  the  ope- 
ration, as  division  is  the  opposite  of  multiplication. 

Obs.  5.  Hence,  to  divide  by  25 — Multiply  the  dividend  by  4  ;  cut 
off  two  figures  from  the  right,  and  tak&  one-fourth  of  these  for  the  true 
remainder. 

18.  Divide  176  by  25.  Ans.  7,  and  1  rem. 
Operation.— \1^  X  4  =  7j04  ;  4  -r-  4  =  1. 

19.  Divide  275  by  25.  Ans.   11. 

20.  Divide  1284  by  25.  Ans.   51,  and  9  rem. 

21.  Divide  4294  by  25.  Ans.   171,  and  19  rem 

22.  Divide  36812  by  25.  Ans.   1472,  and  12  rem. 

Remark. — The  reason  why  we  divide  the  remainder  by  4,  is  because  the  re- 
mainder is  one-?ivndredths,  when  it  should  be  ttcevty -fifths.  We  reduce  it  to 
twenty-fifths  by  dividing  it  by  4.  The  learner  will  understand  this  better 
when  he  has  studied  Fractions.  !!!C!!!^ 

Give  the  remaining  solution  of  this  question.  How  do  we  know  when  onr 
operation  is  completed?  In  examining  this  operation  what  is  the  first  conside- 
ration we  notice?  Tlic  second?  Wliat  is  the  difference  between  this  and  th« 
common  method  of  operating?  Wh»t  ure  .he  chief  adv-sntrg!:"!;  dt-rived  from 
tikis  method?  How  may  we  divide  by  25?  Why  do  we  take  one-feurth  of  tho 
figures  cut  off  for  the  true  remainder? 


Alt.   2.  CONTRACTIONS    AND    ABBREVIATIONS.  65 

Obs.  6.  To  divide  by  125  :  Multiply  the  dividend  hy  8;  cvt  off 
three  figures  from  the  righty  and  take  one-eighth  of  these  for  the  true 
remainder. 

Remark. — The  figures  cut  off  are  thousandths,  whereas  they  should  be 
125ths.     We  reduce  them  to  l'25ths  by  dividing  by  8.     (See  remark  above.) 

23.  Divide  1462  by  125.  Axis.   1 1,  and  87  rem. 
Ojjeration.—\4e2  X  8  =  11|696;  696 -r- 8  =  87. 

24.  Divide  3216  by  125.  Ans.  25,  and  91  rem. 

25.  Divide  4781  by  125.  Ans.  38,  and  31  rem. 

26.  Divide  4821  by  125.  Ans.  38,  and  71  rem. 

27.  Divide  3976  by  125.  Ans.  31,  and  101  rem. 

Obs.  7.  When  there  is  a  remainder  after  dividing  by  several 
numbers,  to  find  the  true  remainder — Midtiply  each  remainder  hy  all 
the  preceding  divisors,  and  to  the  sum  of  their  product  add  the  first  re- 
mainder. 

28.  Suppose  a  teacher  had  158  apples  to  divide  between  6 
classes,  and  each  class  contained  4  scholars.  How  many  should  he 
give  to  eaoh  scholar? 

Ans.  6,  and  he  would  have  14  apples  left,  or  6—  apples  each. 

^         Operation. 
6  158 

26 —  2  rem.    2X6  =  12;   12  +  2  =  14  true  rem. 

6  —  2  rem. 

He  first  divides  the  apples  into  6  piles,  one  for  each  class,  and 
finds  that  each  pile  will  contain  26  apples,  and  there  will  be  2  ap- 
ples left.  He  then  divides  each  pile  into  4  parts,  one  for  each 
scholar,  and  finds  that  there  are  2  apples  left  in  each  pile.  Now  as 
there  are  6  piles,  there  mu^t  be  2  X  6  =  12  apples  left  in  all  the  piles, 
and  the  two  that  were  left  in  the  first  place  make  14,  The  same 
method  of  reasoning  will  apply  to  any  number  of  divisors  ;  hence, 
the  above  rule  is  correet. 


How  do  we  divide  by  125?  Why  do  we  take  one-eighth  of  the  figures  cut 
off  for  the  true  remainder?  When  there  is  a  remainder  after  dividing  by  sev- 
eral divisors,  how  do  we  find  the  true  remainder?  Explain  the  operation  of 
Ex.28,  and  show  why  it  is  correct.  From  what  divisor  is  each  remainder  ex- 
empt from  being  multiphed?  How  may  we  often  contract  operations  in  Long 
Division?  Are  ihe  general  rules,  Sects.  IV.  and  V.,  sufficient  for  all  calcula- 
tions in  Multiplication  and  Division?  Why  then  do  we  use  contractions  and 
abbreviations? 


66  COMMON    ARITHMETIC.  Scct.    VI. 

29.  Divide  2373  by  2,  3,  4,  2  and  3.        Ans.  16,  and  69  rem. 

Operation. 
212373 

Last  rem.  1X2X4X3X2  =  48 

3j  11 86  —  1  rem.  'i'hird  rem,    3X3X2=18 

Second  rem,  1X2  =2 

4|395— Irera.     First  rem,,  1 

2,98  —  3  rem.  Sum  69  true  rem. 

3|49 

16 —  1  rtm. 

Remark. — The  learner  will  perceive  that  no  remainder  is  multiplied  by  th 
divisor  from  which  this  remainder  accrued. 

30.  Divide  1706  by  2,  3, 7,  3,  and  4,       Ans.  3,  and  194  rem. 

31.  Divide  2903  by  3,  4,  2,  and  3.  Ans.  40,  and  23  rem. 

32.  Divide  3721  by  2,  4,  2,  3,  and  2.      Ans.  38,  and  73  rem. 

33.  Divide  4973  by  4,  2,  4,  2,  and  5.    Ans.  15,  and  173  rem. 

34.  Divide  79641  by  4,  7,  6,  8,  9.  and  5.  / 

Ans.   1,  and  19161  rem.   - 
Obs.  8.     We  may  often  contract  operations  in  Long  Division,  by 
rejecting  factors  ;  that  is,  hy  dividing  both  the  dividend  and  divisor  by 
any  number  that  vjill  divide  the  divisor  without  a  remainder. 

35.  Divide  625  by  25.  Ans.  25. 
Operation, 

We  first  divide  both  the  divisorand  dividend  by 

125     5,  and  then  proceed  as  in  Simple  Division. 

Ans.      25 

36.  Divide  34921  by  1728.  Ans.  20,  and  361  rem, 
Opendion. 

8)1728134921  Last  rem.   5  X  8  X  8  =  320 

1 Second  *'     5  X  8  =  40 

8)216i   4365— 1  rem.  First      "  1 


9)27!     545—5    ''  Sum  361  true  rem.   '^ 


3i       60—5 
20 


CANCELATION.  67. 

37.  Divide  4276  by  288.     (288  =  9X8X4.) 

Ans.  14,  and  244  rem. 

38.  Divide  6953  by  256.     (256  =  8X8X4.) 

Ans.  27,  and  41  rem. 

39.  Divide  7491  by  625.      (625  =  5X5X5X5.) 

Ans.   11,  and  616  rem. 

40.  Divide  17426  by  1296.  (1296  =  9X6X8X3.) 

Ans.  13,  and  578  rem. 

Remark. — The  ingenious  pupil,  when  lie  becomes  acquainted  with  Fractions, 
can  easily  study  out  more  contractions,  both  in  Multiplication  and  Division, 
(and  can  derive  both  pleasure  and  profit  from  his  labor);  but  he  niust  recollect 
that  the  General  Rules,  in  Sects.  IV.  and  V.,  will  apply  to  any  case  of  either 
that  may  occur.  The  chief  advantage  derived  from  abbreviations  is,  that  by 
the  aid  of  these  we  can  often  solve  questions  mentally,  which  would  otherwise 
require  the  aid  of  the  slate.  More  methods  might  be  given,  but  is  thought  un- 
necessary, as  these  are  sufficient  for  any  calculations  in  common  business. 


SECTION  VII. 
CANCELATION. 

Article  1.     Definitions  and  Illustrations. 

Obs.  1 .  Cancelation  is  a  short  method  of  i^^rforming  many  op- 
erations of  numbers,  when  Multiplication  and  Division  are  both  con- 
cerned.    To  Ca7icel  means  to  erase,  or  reject. 

Obs.  2.  It  very  often  happens  in  arithmetical  calculations  that 
*we  have,  in  the  same  question,  several  numbers  that  are  multipliers, 
and  several  that  are  divisors.  Now,  if  we  should  so  arrange  these 
numbers  that  all  the  divisors  would  be  by  themselves,  and  all  the  oth- 
er numbers  by  themselves,  it  is  highly  probable  that  the  same  num- 
ber might  occur  both  as  a  multiplier  and  divisor;  or,  if  not,  there 
might  be  factors  common  to  both  multipliers  and  divisors,  which 
might  be  rejected,  and  thus  very  much  shorten  the  operation. — 
This  is  the  object  of  Cancelation. 

Obs.  3.  The  first  thing  we  do  towards  arranging  these  numbers, 
so  as  to  cancel  them,  is  to  draiv  a  perpendicidar  line,  and  place  all 
our  divisors  at  the  left,  and  all  our  other  numbers  at  the  right  hand  side 
of  this  line.     This  is  done  because  it  is  most  convenient. 

Obs.  4.  Now  from  the  very  manner  in  which  our  numbers  are 
placed,  it  follows,  that  the  numbers  at  the  right  are  all  factors  of  some 
number  which  is  a  dividend,  and  the  numbers  on  the  left  are  all  factors 

What  is  Cancelation?  What  does  cancel  mean?  What  often  happens  in 
ariihinetical  calculations?  How  may  we  often  shorten  the  operation  in  such 
cases?     What  is  the  first  thing  we  do  in  order  to  cancel  numbers? 


68 


COMMON    ARITHMETIC, 


•Sect.  YIL 


of  some  number  ivJiich  is  a  divisor,  and  the  ansiver  to  the  cjuestion  is  the 
quotient  of  the  one  divided  by  the  other. 

Therefore,  in  the  following  examples,  we  have  used  the  term 
Dividend  to  represent  the  numbers  at  the  right,  when  taken  collec- 
tively,  and  the  term  Divisor  to  represent  the  numbers  at  the  left, 
when  taken  in  the  same  manner,  and  the  term  Factor  to  represent 
the  numbers  on  either  side  of  the\line  when  taken  separately. 

Ex.  1.  Multiply  24  by  6  and  8,  and  divide  the  result  by  16 
and  6.  Operation. 

We  first  find  that  the  factor  6  is  common     f  —  Xip 
to  both  our  divisor  and  dividend,  and  there-  ^ 

fore  cancel  it.  (Sect.  VI.,  Art.  1,  Obs.  30.) 
We   next  divide  both  our  divisor  and  divi-  ^ins.    1 

dend  by  8;  this  is  done  by  canceling  the  8  in  our  dividend,  into  the 
16  in  our  divisor,  and  setting  the  other  factor  of  16  (2)  on  the  side 
of  our  divisor  (Sect.  VI.,  Art.  1,  Obs.  28.)  Finally,  we  cancel  the 
2  into  24,  which  gives  1 2  as  our  answer. 

2.  Multiply  4667  by  24,  50  and  63,  and  divide  the  result  by  36, 
28,  26  and  40. 


We  first  divide  both  the  divisor 
and  dividend  by  10 ;  this  is  done 
by  canceling  the  cipher  in  40  and 
50.  (Sect,  v..  Art.  4,  Obs.  I.) 
Then  we  divide  both  divisor  and  divi- 
dend by  12;  this  is  done  by  rejecting 
12  in  the  factors  24  and  36,  canceling 
these  factors,  and  placing  each  quo- 
tient (from  dividing  them  by  12,  )on  the 
side  of  the  number  divided  to  produce 
it.  In  the  same  manner  we  cancel  28 
and  63,  dividing  both  by  7.  We  also 
cancel  26  and  4667,  by  dividing  both 
by  13.  The  value  of  the  result  as  yet 
remains  unaltered.  (Sect.  .VI.,  Art.  1, 
Obs.  28.)  We  now  have  a  2  on  each 
cancel,  and  also  cancel  the  3  into  the  9. 


4- 


Operation. 

-n 
■n 


4p 


16 


^4. 
5p 


—  359 


5385(336y'e  Ans 
48 

58 
48 

105 
96 


side  of  the  line,  which  we 
Now  we  have  359,  5,  and 
3  left  on  the  right,  and  two  4's  on  the  left,  and  as  these  will  not  cancel, 
we  multiply  those  on  the  left  together  for"  a  divisor,  and  those  on 
the  right  for  a  dividend,  and  proceed  as  in  Simple  Division. 


Why  are  they  placed  in  this  manner?  What  conclusion  is  deduced  from 
the  numbers  bcinjr  placed  in  this  way?  In  the  examples  given,  what 
do  the  terms  divisor,  dividend  and  factor  represent?  Explain  the  operation 
of  Ex.  1.     Explain  the  operation  of  Ex.  2, 


Art.    1.      ;?,  CANCELATIOh  69 

3.  Multiply  132  by  9,  7,  16  and  12,  and  divide  the  result  by  11, 
144,  63,  8  and  2. 

Operation. 

X\     Xp^  9  times  7  X  63 ;  therefore,   we  cancel  these 

^^  —  XH       '  ^  two  into  63.     8  times  2  =  16;  we  cancel  these 

6yi         /7  into  16.     We  cancel  12  in  144,  which  leaves  12 

'^       X(^  as  the  other  factor  (of  141).     12  times  11  = 

^       X'^  132;  therefore  we  cancel  these  into  132.     Thus 

every  number  cencels,  and  consequently  the  di- 

Ans.  1.  visor  is  equal  to  the  dividend,  and  the  answer  is 
1.     (Sect.  VI.,  Art.  1,  Obs.  26,  d.) 

4.  Multiply  72  by  32,  and  16,  and  divide  the  product  by  144, 
48,  and  24. 

Operation. 


f—XH 
3  —  41^ 
3—^4 

9 


7^  72  cancels  into  144;  16  into  48;  8 

^f  —  4  —  2     divides  24  and  32,  and  2  cancels  into  4. 
X^  Then  we   multiply  the  numbers  on  the 


left  together,  which  gives  9  as  our  di- 
2  =  Ans.  I",  visor,  whilst  our  dividend  is  2.  In  this 
case  our  result  is  fraction,  or  less  than  unity,  (Sect.  YI.,  Art.  1, 
Obs.  26,  d.,)  and  we  can  only  express  the  divisions  by  writing  the 
dividend  over  the  divisor. 

5.  ^Multiply  48  by  24,  and  divide  the  product  by  96  and  72. 
Operation.. 
2 — ^0     4^  In  this  case  all  the  numbers  cancel  on  the 

3  —  ;7^     p\  right  of  the  line  ;  but  when  all  the  numbers 

—  cancel  on  either  side  of  the  line,  the  remain- 

6  \.  Ans.  ing  factor  on  that  side  is  unity,  or  1.  Thus, 
48  in  48  goes  1  time;  48  in  96  goes  2  times;  24  in  24  goes  1  time; 
24  in  72  goes  3  times;  1  X  1  =  1;  2  X  3  =  6;  hence,  we  must  di- 
vide 1  by  6,  and  1  -r-6  =i.  (Sect.  5.  Art.  2.  Obs.  4.  a.) 

Obs.  5.  From  attentively  examining  these  operations,  we  notice 
the  following  considerations  : 

a.  1st.  In  all  questions  in  Cancelation  we  have  an  operating  num- 
ber or  leading  term,  which  is  always  placed  at  the  right. 

b.  2d.  It  does  not  alter  the  result  to  divide  numhers  on  both  sides 
of  the  line  by  the  same  number,  and  canceling  one  number  into  ano- 
ther is  in  reality  dividing  both  sides  by  that  number. 

c.  3d.    When   we  divide  two  factors,  [one  of  the  divisor  and  the 

Explain  the  operation  of  Ex.  3.     Explain  the  operation  of  Ex.  4.     When 

all  the  numbers  cancel  on  both  sides,  what  is  the  result?  When  all  the 
nurnhnrs  cancel  on  either  side  of  the  line  what  is  the  rennaining  factor 
ou  that  silt'?  Uluiitrate  thi^;.  In  exainining  these  operations,  what  is  the 
first  consideration  we  notice?    The  second? 


to*  COMMON    ARITHMETIC.  Scct.    VII 

other  of  the  dividend,^  by  any  number,  we  2^^<^ce  each  residt  on  the 
side  of  the  number  divided  to  produce  it. 

d.  4th.  When  we  have  canceled  all  we  can,  we  multiply  to- 
gether the  remaining  numbers  at  the  left  for  a  divisor,  and  those 
at  the  rigid  for  a  dividend,  and  proceed  as  in  Division  of  Sim- 
ple numbers. 

e.  5 til.  If  all  the  numbers  on  both  sides  of  the  line  cancel,  the 
answer  is  1. 

f  6th.  If,  after  canceling,  the  divisor  is  greater  than  the  dividend, 
the  answer  is  less  than  unity,  and  the  division  can  only  be  expressed. 

g.  7th.  Unity  is  a  factor  on  either  side  of  a  line,  when  all  the 
numbers  cancel  on  that  side. 

Obs.  6.  The  learner  must  bear  in  mind,  in  all  his  operations, 
that  the  principle  of  cancelation  co7isists,  siniply,  in  rejecting  the  fac- 
tors common  to  both  the  divisor  and  dividend.  This  being  the  case, 
all  our  canceling  is  performed  by  Division.  Some,  however,  attempt 
to  explain  it  on  the  principles  of  Subtraction,  saying,  that  if  equals 
a.re  taken  from  equals,  the  remainder  will  be  equal.  The  principle  is 
correct ;  but  as  our  divisor  and  dividend  are  not  always  equal,  un- 
fortunately, this  principle  cannot  apply  to  Cancelation  ;  because  the 
subtraction  of  the  same  number  from  unequal  quantities,  where  one 
is  a  divisor  and  the  other  a  dividend,  will  materially  alter  the  value 
of  the  quotient.    (Sect.  VI.,  Art.  1,  Obs.  29.) 

From  the  preceding  remarks  and  illustrations,  we  derive  the  fol- 
lowing 

GENERAL  RULE  FOR  CANCELATION. 

I.  Draw  a  perpendicular  line,  and  place  the  operating  number.,  or 
leading  term,  at  the  right,  together  with  all  the  midtip>liers,  and  all\the 
other  numbers  at  the  left.     (Obs.  3  and  5,  a.) 

II.  When  the  sa7ne  factor  occurs  on  both  sides  of  the  line,  cancel 
it  in  both  places. 

in.  After  canceling  all  common  factors,  multiply  together  all  the 
remaining  numbers  at  the  right  for  a  dividend,  and  those  at  the  left  for 
a  divisor,  and  thenp>roceed  as  in  Division  of  Simple  numbers.  (Obs. 
5,d.) 

Note. — Read  carefully  the  considerations  under  Obs.  5. 


Third?  Fourth?  Fifth?  Sixth?  Seventh?  In  what  does  the  principle 
of  Cancelation  consist?  How  is  it  performed?  How  do  some  explain  it? 
Why  do  they  explain  it  thus?  Is  this  principle  correct?  Why,  then,  will  it 
not  apply  to  Cancelation?  Will  it  produce  the  same  final  result  to  subtract 
the  sania  number  from  unequal  quantities,  when  one  is  a  divisor  and  the  other 
a  dividend?  Why  not?  In  the  General  Rule,  what  is  the  first  thing  we  do? 
What  is  the  second?    The  third?     •    v^a™  «^:i     f.-i^iv^z  ■i- •  f..*?j*s5«,rj:*5n«r7  jins 


Art.    1.  CANCELATION.  71 


EXERCISES    FOR    THE    SLATE. 


1.  If  6  bushels  of  potatoes  cost  18  shillings,  how  much  would  9 
bushels  cost? 


Operation. 

Mrs  — 3 
9 


If  6  bushels  cost  18  shillings,  1  bushel 
will  cost  18  -f-  6=3  shilllings,  and  9  bushels 
would  cost  9  times  as  much  as  1  bushel. — 
Hence,  in  this  example  we  divide  by  6  and  Ans.  27  shillings. 

multiply  by  9. 

2.  If  16  yards  of  cloth  cost  64  dollars,  how  much  would  19 
yards  cost?  Ans.  76  dollars. 

3.  Paid  50  dollars  for  16  sheep .  how  much  should  I  pay  for  40 
sheep  at  the  same  rate?  Ans.   125  dollars. 

4.  How  much  must  I  pay  for  36  horses,  if  I  pay  450  dollars  for 
11  horses?  "  Ans.   1472yV  dollars. 

5.  How  much  must  I  pay  for  32  yards  of  broadcloth,  if  8  yards 
cost  48  dollars?  Ans.   192  dollars. 

6.  How  much  must  I  pay  for  18  cows,  if  6  cows  cost  90  dollars? 

Ans.  270  dollars. 

7.  There  are  365  days  in  a  year,  and  24  hours  in  a  day,  and  the 
earth  moves  around  the  sun  at  the  rate  of  68000  miles  an  hour. 
Now  how  many  years  would  it  take  a  man  to  travel  the  distance  the 
earth  moves  in  a  year,  allowing  a  man  to  travel  40  miles  a  day? 

Ans.  40800. 

8.  How  long  would  it  take  him,  if  he  traveled  60  miles  a  day? 

Ans.  272000  years. 

9.  How  long,  if  he  traveled  80  miles  a  day? 

^*  Ans.  20400  years. 

10.  If  a  man  walk  200  miles  in  6  days,  how  many  miles  can  he 
walk  in  28  days?  Ans.  933^. 

11.  If  I  pay  24  dollars  for  12  books,  how  many  dollars  must  I 
pay  for  19  books?  Ans.  38. 

1 2.  How  much  would  40  acres  of  land  cost,  if  30  acres  cost 
150  dollars?  Ans.  200  dollars. 

13.  How  much  would  60  bushels  of  wheat  cost,  if  25  bushels 
cost  30  dollars?  Ans.  72  dollars. 

14.  How  much  would  76  bushels  of  oats  cost,  if  15  bushels  cost  4 
dollars?  Ans.  20yV  dollar. 

15.  If  I  pay  150  dollars  for  6  stoves,  how  much  must  I  pay  for 
14  stoves?  Ans.  350  dollars. 

16.  18  times  432  are  how  many  times  12.         Ans.  648  times. 

17.  Multiply  847  by  19,  28,  and  54,  and  divide  the  result  by  57, 
36,  and  98.  Ans.   121. 

18.  Multiply  8700  by  91,  46,  and  144,  and  divide  the  result  by 
390,  132,  and  1740.  Ans.  58y\. 


1^  COMMON    ARITHMETIC.  Scct.    VII. 

19.  Multiply  50  by  128,  75,  35,  and  85,  and  divide  the  result  by 
1200,  1360,  and  875.  Ans.   1. 

20.  Multiply  380  by  57,  115,  186,  and  323,  and  divide  the  result 
by  760,   171,  92,  589,  and  765.  Ans.  3^- 

Article  2.     Greatest  Common  Divisor. 

Suppose  it  were  required  to  find  some  number  that  would  divide 
8  and  12  without  a  remainder.  We  find  that  2  will  divide  both, 
because  8-^2  =  4,  and  12  -v- 2  =  6.  Then  2  is  a  common  divi- 
sor of  8  and  12.     Hence — 

Obs.  1,  A  common  divisor  of  two  or  more  numbers,  is  a  num- 
her  that  will  divide  them  xiyiihout  a  remainder. 

But  suppose  we  wished  to  find  the  greatest  number  that  would 
divide  8  and  12.  This  we  find  to  be  4,  as  8  -r-  4  =  2,  and  12  -f-  4 
=  3,  and  2  and  3  are  both  prime  numbers,  and  cannot  be  divided 
ao^ain.  Then  4  is  the  greatest  common  divisor  of  8  and  12. — 
Hence — 

Obs.  3.  The  greatest  Common  Divisor  of  two  or  more  num- 
hers   is  the  greatest  number  that  will  divide  them  ivithout  a  remainder. 

Remark  1. — One  number  is  said  to  be  a  measure  of  anotlipr  number,  when 
ihQ  former  is  contained  in  the  latter  without  a  remainder.  Therefor**,  a  com- 
mon divisor  is  often  called  a  Common  Measure  ;  and  the  greatest  common 
divisor  of  two  or  more  numbers,  is  often  called  the  Greatest  Common  Mea- 
sure of  those  numbers. 

2.  It  must  be  observed,  however,  that  whilst  a  measure  can  exist  with  re- 
ference to  two  numbers,  the  common  measure,  and  the  greatest  common 
measure  must  both  always  have  reference  to  at  least  three,  one  of  which  must 
divide  the  other  two.  Thus,  4  is  a  measure  of  16,^a  common  measure  of  20 
and  24,  and  the  greatest  common  measure  of  8  and  12. 

Ex.  1.  What  is  the  greatest  common  divisor  of  9  and  126? 
Solution. — The  greatest  common  divisor  cannot  be  greater  than 
9,  because  it  must  divide  9.     9  will  divide  itself;  let  us  see  if  it 
will  divide  126.     We  find  by  trial,   that  126  contains  9  exactly   14 
times  ;  then  9  must  be  the  greatest  common  divisor  of  9  and  126. 
2.  What  is  the  greatest  common  divisor  of  15  and  80? 
Operation. 

15)80(8  We  first  try  if  15  is  the  greatest  com- 

75  mon  divisor.     After  dividing  80  by  15, 

—  we   have   a  remainder  of  5.     Now   if  5 

5)15(3       will  divide  15,  it  will  also  divide  80  ;  be- 

1 5  cause  if  1 5  contains   5,   75  will  also  con- 

—  tain  5,  as  75  is  divisible  by  1 5  ;  then  if 

00  75  contains  5,  80  will  also  contain  5 ;  be- 


What  is  a  common  divisor  of  2  or  more  numbers?  The  greatest  common 
divisor?  When  is  one  number  said  to  be  a  measure  of  another?  What  is  a 
common  divisor  often  called?  The  greatest  common  divisor?  What  is  the 
difference  between  a  measure,  a  common^measure,  and  the  greatest  common 
measure  of  numbers?    Give  an  example.'  Explain  the  solution  of  Ex.  1. 


Art.   2.  GREATEST   COMMON   DIVISOR.  T3 

cause  80  =  75  +  5  ;  that  is,  80  will  contain  5,  one  more  time  than 
75.  We  find  that  15  will  contain  5,  therefore  5  is  the  greatest  com- 
mon divisor  of  15  and  80. 

From  this  example,  we  perceive  that  the  greatest  common  di- 
visor of  two  numbers,  must  be  a  common  divisor  of  the  least  number 
and  their  remainder  after  division  ;  therefore,  it  cannot  be  larger  than 
this  remainder,  because  the  remainder  must  be  divided  by  it. — 
Hence — 

To  find  the  greatest  common  divisor  of  two  numbers,  we  have, 
the  following 

Rule. — Divide  the  greater  number  by  the  less,  and  thai  divisor  by 
the  remainder,  and  so  on,  continuing  to  divide  the  last  divisor  by  tlie  last 
remainder,  till  nothing  remains.  The  last  divisor  will  be  the  greatest 
common  divisor. 

Note. — 1.  The  learner  will  perceive  that  a  common  dirisor  of  two  or  mora 
numbers,  is  simply  a  common  factor  of  those  numbers  ;  and  the  greatest  com- 
mon divisor  is  iheir  greateit  common  factor. 

ij  2.  We  learn,  (Sect.  IV.  Art.  4,  Obs.  3,  Rem.  I.')  that  a  prime  number  cannot 
consist  of  two  factors,  both  greater  than  unity,  or  1.  Therefore,  a  prime  num' 
her  is  divisible  only  by  itself  and  unity. 

3.  One  number  is  said  to  he  prime  to  another,  when  only  a  unit  will  divido 
both  of  them.    Hence — 

Obs.  3.  Two  or  more  prime  numbers  cannot  have  a  common  divi- 
sor, because  they  have  nx)  common  factors. 

EXERCISES    FOR    THE    SLATE. 

1.  What  is  the  greatest  common  divisor  of  18  and  78? 

Ans,  6. 

2.  What  is  the  greatest  common  divisor  of  23  and  120? 

Ans,  1. 

3.  What  is  the  greatest  common  divisor  of  32  and  122? 

Ans.  2. 

4.  What  is  the  greatest  common  divisor  of  69  and  291?. 

Ans,  3. 

5.  What  is  the  greatest  common  divisor  of  231  and  517? 

Ans.   11. 

6.  What  is  the  greatest  common  divisor  of  5191  and  8497? 

Ans.  29. 
To  find  the  greatest  common  divisor  of  more  than  two  numbers  : 
Obs.  4.     Find  the  greatest  common  divisor  of  two  of  them;  then 

Explain  the  operation  of  Ex.  2.  What  consideration  do  we  notice  in  the  so- 
lution of  this  example?  How  do  we  find  the  greatest  common  divisor  of  two 
numbers?  What  is  a  common  divisor?  The  greatest  common  divisor?  Of 
what  cannot  a  prime  number  consist?  What  inference  do  we  deduce  from 
this?  When  is  a  number  said  to  be  prime  to  another?  Can  two  or  more 
prime  numbers  have  a  common  divisor?  Why  not?  What  is  the  first  rule 
for  finding  the  greatest  common  divisor  of  more  than  two  numbers? 

5 


74  CJOMMON   ARITHSIETia  Scct.  "Vil. 

of  this  greatest  common  divisor,  and  another  given  number,  and  so  on, 
through  all  the  given  numbers.  The  last  common  divisor  found  will 
be  the  one  required.     Or, 

Obs.  5.  W7'ite  the  numbers  in  a  line.  Divide  them  by  any  num- 
ber that  will  divide  them  all  without  a  remainder,  and  write  the  quotients 
hdow,  as  in  Division. 

Divide  these  quotients  in  the  same  manner,  and  so  continue  to  do, 
till  no  number  greater  than  1  will  divide  them  all  without  a  remainder. 

Finally,  multiply  together  all  the  divisors,  and  their  product  will 
he  the  greatest  common  divisor. 

Demonstration. — ^We  learn,  Sect.  lY,  Art.  4,  Obs.  2,  that  every 
composite  number  is  composed  of  factors;  and  if  we  resolve  two  or 
more  numbers  into  their  several  factors,  and  the  same  factor  occurs 
in  all  the  numbers,  this  factor  is  evidently  a  common  divisor  of  these 
numbers,  and  \he  product  of  all  these  common  factors  is  \he\r  great- 
est common  divisor.  Now  these  common  factors  are  the  numbers  by 
which  we  divide;  hence,  the  above  rule  is  correct. 

7.  What  is  the  greatest  common  divisor  of  64  and  96? 

Operation. 

8)64.  .96         We  first  divide  by  8  and  then  by  4.  2  and  3  are 

not  divisible  by  any  number  greater  than    1,  hence, 

4)8 12     8  X  4  =  32  the  greatest  common  divisor  required. 

2._  3 
8X4=32  Ans. 

8.  What  is  the  greatest  common  divisor  of  2457,  3213,  and 
1197? 

By  the  first  Iltde.  By  the  seconid  Rule. 


2457)3213(1 
2457 

189)1197(6 

1  1S4. 

9 

2457.3213_1197 

7 

273..357__133 

756)2457(3 
2268 

Ans.  63)189(3 
189 

39  ..  51  ._  19 

189)756(4 
756 

000 

9  X  7  ==  63.  Ans. 

000 

Obs.  6.     The  following  rules  may  assist  the  learner  in   finding 
eommon  divisors : 
a.     Any  even  number  may  be  divided  by  2,  because  as  the  remain- 

The  iecond  rule?    Demonstrate  this  rule.    What  numbers  can  be  divided 
^ya?    By  3?    By  9? 


Art.    2,'^^^  GREATEST   COMMON   riVISOR.  75 

der  must  always  be  less  than  the  divisor,  (Sect.  V,  Art.  2,  Obs.  12, 
b.)  the  last  partial  dividend  must  be  0,  2,  4,  G,  ft,  10,  12,  16,  or 
18,  either  of  which  will  contain  2. 

b.  Any  number  can  be  divided  by  3  when  the  sum  of  its  dibits  can 
be  divided  by  3. 

c.  Also,  Any  maiTher  can  be  divided  by  9  when  the  sum  of  its  dig- 
its can  be  divided  by  9. 

The  last  two  rules  are  both  demonstrated  in  the  same  manner, 
(Sect  VI,  Art.  1,  Obs.  17,  demonstration.) 

d.  Any  number  can  be  divided  by  4,  when  ils  two  right  hand  digits 
can  be  divided  by  4.  Because,  if  the  number  can  be  divided  by  4,  it 
can  be  divided  by  2  X  2  ;  (Sect.  V.,  Art.  4,  Obs.  2,  a.)  there- 
fore, the  half  of  such  a  number  must  be  an  even  numbc  t}  and 
if  half  of  any  number  will  contain  2,  the  number  itself  wiil  con- 
tain 4. 

e.  Any  number  can  be  divided  by  5,  ivhen  its  right  hand  figure  is  5 
or  0.  Because,  as  the  remainder  is  always  less  than  the  divisor, 
(Sect,  v..  Art.  2,  Obs  12,  b.,)  the  last  partial  dividend  must  be  5, 
10,  15,  20,  25,  30,  35,  40,  or  45,  all  of  which  will  contain  5. 

f.  Any  nuriiher  ending  with  0,  00,  000,  &c.,  can  be  divided  by  10, 
too,  1000,  &c.     (Sect  v.,  Art.  4,  Case  1.) 

g.  Xo  even  number  will  divide  an  odd  number,  and  no  number 
Will  contain  a  greater  number  than  its  half.     Because — 

1st.  If  an  odd  number  will  contain  an  even  number,  it  will  also 
contain  2,  which  is  a  factor  of  all  even  numbers  ;  (Sect.  V.,  Art.  4, 
Obs.  2,  a.)  But  this  cannot  be  according  to  Sect.  IV.,  Art.  4, 
Obs.  3,  Rem.  3.  The  reason  why  an  odd  number  will  not  contain 
2,  is  because  the  last  partial  dividend  is  always  1,  3,  5,  7,  9,  11, 
13,  15,  17,  or  19,  none  of  which  will  contain  2  without  a  re- 
mainder. 

2nd.  Any  number  will  contain  its  half  just  twice.  Hence,  if  the 
divisor  is  greater  than  the  half  of  the  number,  it  must  be  equal  to 
the  number  itself,  or  there  will  be  a  remainder  after  the  division  is 
performed  ;  else  it  must  be  greater  than  the  number,  when  the 
result  is  a  fraction.  (Sect.  VI.,  Art.  1,  Obs.  26,  d.) 

Find  the  greatest  common  divisor  of  the  following  numbers : 

9.  48  and  60.  Ans.  12. 

10.  24,  36,  and  84.  Ans.   12. 

11.  36,  96,  and  144.  Ans.   12. 

12.  25,  45,  85,  and  115.  Ans.  5. 

13.  22,  44,  143,  209,  and  297.  Ans.  11. 

14.  18,  63,  99,vX17,  ajfld  171.  Ans.  9. 

By  4?  By  5?  By  10,  100,  1000,  &c.?  Will  an  even  number  divide 
an  odd  number?  Why  not?  What  is  the  greatest  divisor  any  number  will 
contalo?     Demonstrate  these  rules. 


Sect, 

.VIL 

Ans, 

.  3. 

Ans. 

83. 

Ans. 

47. 

Ans. 

13. 

Ans 

.  4. 

Ans. 

.7. 

76  COMMON  arithmetic! 

15.  48,  54,  75,  and  111. 

16.  249,  332,  and  415. 

17.  94,  188,  282,  and  423. 

18.  78,  117,  143,  and  169. 

19.  12,  20,  36,44,  and  48. 

20.  21,  49,  63,  91,  and  133. 

Article  3.     Least  Common  Multipli^ 

Obs.  1 .  One  number  is  said  to  be  a  Multiple  of  another  number, 
when  the  former  contains  the  latter  tvithout  a  remainder.  Thus,  4  or  6 
is  a  multiple  of  2,  because  either  of  them  will  contain  2. 

Obs.  2.  A  Common  Multiple  is  amj  number  that  will  contain  two 
or  more  numbers  without  a  remainder.  Thus,  1 2  is  a  common  multi- 
ple of  3  and  4,  because  it  will  contain  both  3  and  4  without  a  re- 
mainder. 

Remark. — A  multiple,  whether  common  of  <m«,  two,  or  more  numbers,  t» 
always  a  composite  number,  and  the  numbers  contained  in  it  are  its  factors. 
Thus,  2  is  a  factor  of  4,  and  3  and  4  are  factors  of  12. 

Hence — To  find  a  common  multiple  of  two  or  more  numbers  :    '' 
Obs.  3.     Multiply  them  together. 

Ex.  1.  What  is  a  common  multiple  of  3,  4,  and  6.     Ans.  72. 

2.  Of  4,  8,  and  10.  Ans.  320. 

3.  Of  2,  4,  6,  8,  10,  and  12.  Ans.  46080. 

Remark  1. — It  is  sometimes  desirable  to  find  the  least  number  that  will  con- 
tain two  or  more  numbers  without  a  remainder.  This  is  called  their  Least 
Common  Multiple.  Thus,  72  is  a  common  multiple  of  3,  4,  and  G  ;  but  12  is 
their  least  common  multiple. 

2  A  common  multiple  of  two  or  more  numbers  must  contain  all  the  differ- 
ent factors  of  these  numbers.  Thus,  48,  'a  common  multiple  of  6  and  8,  con- 
tains|3  and  2,  the  factors  of  6,  and  also  4  and  2,  or  2,  2,  and  2,  the  factors  of 
8  ;  but  32  will  not  contain  the  factors  of  6  and  8,  therefore  it  is  not  a  common 
multiple  of  these  numbers, 

4.  What  is  the  least  common  multiple  of  3,  5,  and  7? 

These  numbers  are  sill  prime,  and  have  no  factors  ;  therefore,  no 
number  less  than  their  continued  product  will  contain  them  ;  that  is 
3  X  5  X  7  ==  105.  Ans.     Hence — 

To  find  the  least  common  multiple  of  prime  numbers — 

Obs.  4.     Multiply  them  together. 

When  is  one  number  said  to  be  a  multiple  of  another?  Give  an  example. 
What  is  a  common  multiple?  Giv«  an  example.  What  kind  of  a  rumber  is 
a  multiple?  What  are  its  factors?  Give  examples.  How  do  we  find  a  com- 
mon multiple,  or  two  or  more  numbers?  What  is  the  least  common  multiple 
of  two  or  more  numbers?  Give  an  example.  What  must  all  common  multi- 
plea  of  two  or  more  numbers  contain?  Give  an  example.  What  factor  of  6 
or  8  will  not  32  contain? 


Art.    3.  LEAST   COMMON   MULTIPLE.  77 

6.  What  is  the  least  common  multiple  of  2,  3,  and  5? 

Ans.  30. 

6.  What  is  the  least  common  multiple  of  7,  11,  and  13? 

Ans.   1001. 

7.  What  is  the  least  common  multiple  of  17,  19,  23,  and  29? 

Ans.  215441. 
To  find  the  least  common  multiple  of  composite  numbers  : 
Obs.  5.     As  we  have  said  before,    (Sect.  V.,  Art.  4,  Obs.  2,  a,) 
if  any  number  will  contain  a  composite  number,  it  will  also  contain  its 
fojctors.     Thus,  24  will  contain  12,  and  it  will  also  conta'n  2  and  6, 
or  3  and  4,  the  factors  of  12. 

8.  What  is  the  least  common  multiple  of  6  and  8? 

Solution, — The  factors  of  6  are  2  X  3  ;  of  8,  2  X  2  X  2.  IS'ow 
we  wish  our  multiple  to  contain  all  these  factors,  but  it  is  evident 
that  if  the  same  factor  occurs  in  both  numbers  it  will  occur  twice  in 
their  product,  or  common  multiple,  where  it  need  only  occur  once. 
Then  as  the  factor  2  occurs  in  both  6  and  8,  we  will  reject  it  in  one 
of  tliem,  and  the  product  of  the  other  factors  3X2X2X2  =  24, 
the  least  common  multiple  required. 

9.  What  is  the  least  common  multiple  of  2,  4,  6,  8,  12,  and  16? 
Operation. 

^-_^_.^_-^-_^^-_16  Any  number  that  we  can  divide  by 

f;  3  16,  we  can  divide  by  2,  4,  and  8,  I  e- 

fi  16  X  3=  48  Ans.  cause  these  numbers  are  factors  of  16; 

/  also,  any  number  that  we  can  divide  by  12,  we  can  divide  by  6,  as 

6  is  a  factor  of  12.  (Obs.  5.)     Now  we  have  remaining  12  and  16, 

but  these  numbers  have  a  common  factor,  (4.)     This  we  reject  in 

the  12,  and  multiplying  16  by  the  other  factor  of  12,  (3,)  we  obtain 

48  as  our  answer. 

Remark. — We  write  our  numbers,  placing  the  largest  at  the  right,  because 
it  is  more  convenient  to  place  them  in  this  manner. 

t*       10.  What  is  the  least  common  multiple  of  3,  9,  8,  15,  6,  5,  14, 
-10  and  12? 

Operation. 
1st  line,  ^__^__^..?..^--J'0..^^._14_.15 
2nd  "  3__  ^__  ^ 

3rd   "  ^         .-         2 

4th    "  2 

2X2X3X14X15  =  2520  Ans. 

When  any  number  will  contain  a  composite  number,  what  must  it  also  con- 
tain? Give  an  example.  Explain  the  solution  of  Ex.8.  Why  do  we  reject 
the  factor  2  from  one  of  the  numbers  in  this  example.  Explain  the  operation 
of  Ex.  9.  How  do  we  write  our  numbers?  Why?  Explain  the  operation  of 
Ex.10. 


78  COMMON    ARITHMETia  Scct.    VIL 

We  first  cancel  3,  5,  and  6,  because  tliey  are  factors  of  9,  10, 
and  12.  We  next  find  that  the  factor  3  is  common  to  9  and  15; 
we  cancel  it  in  the  9  ;  5  is  a  common  factor  of  10  and  15  ;  we  can- 
cel it  in  the  10  ;  3  is  a  common  factor  of  12  and  15  ;  we  cancel  it 
in  the  12  ;  the  remaining  factors  of  these  numbers  (9,  10,  and  12,) 
we  write  below,  and  these  (the  factors  3,  2,  and  4),  form  the 
second  line.  Next,  we  find  that  the  factor  2  is  common  to  8  and 
14;  we  cancel  it  in  the  8;  2  (in  the  2nd  line,)  will  divide  14 ;  we 
cancel  it,  (2;)  2  is  a  factor  common  to  4,  (in  the  2nd  line,)  and 
14  ;  we  cancel  it  in  the  4  ;  we  write  the  remaining  factors  (of  8  and 
4)  below,  as  our  third  line.  Next,  we  find  the  factor  2  common  to 
4  and  2,  (in  the  third  line  ;)  rejecting  it  in  the  4,  we  write  the  re- 
maining factor  (of  4)  below  as  our  fourth  line.  Finally,  we  multi- 
ply together  the  numbers  remaining  (in  all  the  lines,)  and  their  pro- 
duct is  the  least  common  multiple  required. 

Obs.  6.  From  examining  this  operation,  we  notice  the  following 
considerations  : 

1st.  We  work,  in  all  cases,  from  the  left  hand  towards  the  rtght, 
because  it  is  more  convenient,  from  the  manner  in  which  the  num- 
bers are  written.     Thus  : 

a.  We  cancel  a  numher  at  the  left ,  ivheni  it  will  divide  one  at  the 
right,  because  the  number  at  the  left  is  ?i  factor  of  the  number  at  the 
right. 

b.  When  we  cancel  a  common  factor,  we  cancel  it  in  the  nvmher  at  the 
left.  This  is  done,  because  the  number  at  the  left  being  smaller  than 
the  number  at  the  right,  the  factor  that  is  not  canceled  (at  the  left) 
is  smaller  than  the  corresponding  factor  of  the  number  at  the  right, 
and  by  this  means  we  obtain  the  smallest  numbers  as  our  final  mul- 
tipliers. 

2d.  When  we  reject  common  factors  from  two  niirrtbers,  we  reject  the 
GREATEST  FACTOR  comvtion  to  them. 

This  is  done  to  prevent  the  same  factor  ivova  occwvr'mg  more  times 
than  is  necessary  in  the  multiple.  This  would  frequently  be  the 
case  if  we  simply  canceled  the  smaller  factors.  Besides,  by  cancel- 
ing the  greater  factors,  our  final  multipliers  are  smaller. 

3d.  We  cannot  i^eject  factors  twice  from  the  same  numbers.  Be- 
cause the  first  time,  we  cancel  the  greatest  factor  common,  and  if  we 
should  again  cancel  factors  from  the  same  numbers,  our  multiple 
might  not  contain  both  these  numbers.  Thus,  (in  the  last  exam- 
ple) after  canceling  the  factor  3  from  9,  (because  3  is  a  common  fac- 


n  we  notice  from  exsimining  this  operation?  Why? 
first  place.     Why?     How  iu  the  second   phice? 
uu  WB  uaucui  Luu  ]acLor  iathe  number  at  the    left?     What  is  the  second 
deration  wc  notice?     Why  do  wo  reject   the  greatest  factor?     What  is 
the  third  cousideration?    Wliy  not? 


Wiiat  is  the  first  consideration 
Show  how  v/c  do  this  in  the 
Why  do  we  cancel  the  facto 


Art.    3.        •  LEAST    COMMON    MULTIPLE.  79 

tor  of  9  and  15,)  some  might  say  that  3  (the  other  fa«tor  of  9)  was 
a  factor  of  1 5,  and  cancel  it  accordingly.  This  would,  in  reality,  be 
canceling  9  into  15,  or  making  9  a  factor  of  15,  which  is  not  the 
case,  and  the  consequence  would  be  that  the  final  result  would  not 
contain  9. 

4th.  We  first  cancel  factors  with  reference  to  the  right  hand  num- 
ber; next,  with  reference  to  the  second  number  from  the  right,  and  then 
with  reference  to  the  third  number  frcmi  the  rights  and  so  on.  We  do 
this,  because  the  numbers  at  the  right  being  the  largest,  it  is  highly 
probable  that  many  of  the  smaller  numbers  will  cancel  out  entirely 
in  these,  and  thus  we  would  have  fewer  numbers  to  multiply  to- 
gether to  obtain  our  final  result. 

5th.  After  canceling  all  we  can,  we  multiply  the  remaining  numbers 
together.  We  do  this,  because,  after  all  the  common  factors  are  can- 
celed, the  least  common  multiple  must  be  the  product  of  the  remain- 
ing numbers. 

Some  may  think  that  we  can  reject  factors  indiscriminately  and 
obtain  the  same  result.  The  following  example  will  show  their 
error  : 

11.  What  is  the  least  common  multiple  of  3,  9,  12,  16, 18,  and 
24? 

First  Operation.  Second  Operation.  Third  Operation. 

^.^.X$.X^M-^4r  X^.^-X^-^.U,X$  ^.^.X^.X^.X^.24 

2_  3  ^3  rf-6 

2 
24X2X3=144  24X3=72  24X6X2=288. 

an  incorrect  result.  an  incorrect  result 

In  the  second  operation  we  canceled  2  into  the  1 6,  (it  being  com- 
mon to  16  and  18  ;)  and  6  mto  the  18,  (it  being  common  to  18  and 
24  ;)  and  8  into  24.     Then  24  X  3  =  72,  a  result  too  small. 

In  the  third  operation  we  did  not  cancel  the  greatest  common  fac- 
tor each  time,  and  the  result  was  too  large. 

To  show  that  the  first  result  is  correct,  we  will  resolve  144  into 
its  several  factors,  and  see  if  all  the  different  factors  of  the  given 
numbers  occur  in  it. 

The  factors  of  144  are  3  X  3  X  2  X  2  X  2  X  2. 

The  factors  of  the  given  numbers  are3X   3X3X2X2X3X 


2X2X2X2  X  2X3X3  X  2X2X2X3. 


Gire  an  example  illustratinfr  this,  and  show  the  consequence  upon  the  re- 
sult by  departing:  from  this  rule?  What  is  the  fourth  consideration?  Why  do 
we  do  this?  Why  will  it  notdo  to  reject  factors  indiscriminately?  (Explain  a 
few  examples  on'lhe  black-board  illustrating  this  point.)  Htw  do  we  show 
that  the  result  ef  the  first  operation,  Ex.  11,  waa  correct? 


80  Common  aeithmetic.  Sect.  VII. 

I    Canceling  superfluous  factors ^Xy^X^X^X^Xy^X 

2X2X2X2X^X3  XSX^X^X^X^ 

We  have  2X2  X2X2X3X3=  144.     Therefore,  it  is  cor- 
rect. 
I    The  braces  include  the  factors  of  each  number. 

Hence — To  find  the  least  common  multiple  of  two  or  more  num- 
bers, we  have  the  following 

RtTLE. — I.  Set  the  numbers  down  in  a  line,  placing  the  greatest  at 
the  right.     (Rem.  3.) 

II.  When  the  numbers  at  the  left  have  factors  common  to  the  right 
hand  number,  cancel  the  factors  at  the  left  only;  so  proceed  till  all  the 
cxmimon  factors  at  the  left  are  canceled. 

III.  Multiply  together  the  remaining  factors,  and  their  product  will  he 
the  least  common  multiple.     (Obs.  6,  6th.) 

Note. — The  learner  will  perceive  that  when  the  second  and  third  numbers 
from  the  right  cancel  in  the  first  part  of  the  operation,  we  take  their  factors 
ieloto,  and  proceed  as  with  the  numbers  themselves.  No  numbers  are  used  af- 
ter being  canceled,  as  all  their  factors  are  found  in  the  other  numbers. 

EXERCISES   FOR   THE    SLATE. 

What  is  the  least  common  multiple  of  the  follov^ing  numbers  : 

1.  6,  8,  Ans.  24. 

2.  7,  12,  Ans.  84. 
|r    3.  9,  12,  18,  Ans.  36. 

4.  8,  15,24,               '  "vSj  Ans.  120. 

5.  6,  9,  15,  20,            -^'^  '                     Ans.   180. 

6.  6,  7,  10,  25,  Ans.  350. 

7.  10,  12,  16,  18,  24,  V               Ans.  720. 

8.  10,  16,  20,  24,  32,  Ans.  480. 

9.  6,  8,  10,  16,  18,  20,  Ans.   720. 

10.  3,  5,  2,  9,  12,  15,  18,  Ans.   180. 

11.  3,  4,  8,  12,  16,  20,  24,  28,  Ans.   1680. 

12.  4,  2,  6,  8,  9,  12,  14,  15,  16,  Ans.  5040. 

13.  6,  2,  36,  10,  12,  18,  20,  24,  Ans.  360. 

14.  6,  4,  8,  12,  16,   24,  32,  Ans.  96. 

15.  1,  2,  3,  4,  5,  6,  7,  8,  9,  10,  Ans.  2520. 

16.  8,  12,  14,  18,  24,  32,  36,  40,  Ans.   10080. 

17.  9,  12,  16,  32,  36,  48,  96,  Ans.  288. 

18.  10,  3,  8,  12,  32,  16,  27,  18,  Ans.  4320. 

19.  7,  11,  22,  49,  55,  70,  84,  Ans.  32340. 

20.  12,  8,  14,  18,  24,  15,  36,  28,  Ans.  2520. 

What  is  the  first  step  in  finding  the  least  common  multiple  of  numbers? — 
The  second?  Third?  When  the  second  and  third  numbers  from  the  right 
cancel  how  do  we  proceed?  Do  we  use  a  number  after  it  is  canceled?  Why 
not? 


FRACTIONS.  <81 

SECTION-  VIII. 
FRACTIONS. 

Article  1.     Mental  Exercises,  Definitions,  &c. 

Obs.  1.     A  Fraction  is  apart  of  any  ildng. 

Thus  if  I  speak  of  an  apple,  I  evidently  mean  a  whole  apple  ; 
but  if  I  cut  the  apple  miotwo  equal  parts,  each  part  is  called  o?ie- 
half ;  if  I  cut  it  into  three  equal  parts,  each  part  is  called  one-third; 
if  I  cut  it  into  Jour  equal  parts,  each  part  is  calkd  one-fourth,  and 
two  of  these  last  named  parts  are  called  two-fourths;  three  parts 
are;  called  three  fourths,  &c.,  and  the  expressions,  one-half,  one- 
third,  one-fourth,  two-fourths,  three-fourths,  &c.,  are  called  Frac- 
tions. 

We  can  divide  numbers  into  parts  as  well  as  apples;  thus  we  can 
take  one-half  of  2,  one-third  of  3,  one-fourth  of  8,  or  any  other  part 
of  any  number. 

To  take  one-half  of  any  number,  we  divide  it  by  2 ;  to  take  one- 
third  of  any  number,  we  divide  it  by  3  ;  to  take  one-fourth  of  any 
number,  we  divide  it  by  4,  &c. 

Ex.  1.  What  is  one-half  of  the  following'numbers?  4,  6,  8, 
10,   12,   14,  18,    16,  24,  22,   20,  36. 

2.'^  i,What  is  one-third  of  the  following  numbers  ?  6,  9,  12,  15, 
18,  24,  21,  36,  33,   30,  27. 

3.  What  is  one-fourth  of  the  following  numbers?  4,  8,12,16, 
24,  20,  36,  32,  28,  40,  44. 

4.  What  is  one-fifth  of  the  following  numbers?     5,   10,   15,  25, 

20,  45,  30,  40,  36,  60,   50,  45,  55. 

5.  What  is  one-sixth  of  the  following  -  numbers?  12,  18,  48, 
24,  36,  30,  42,  60,  QQ. 

6.  What  is  one-seventh  of  the  following  numbers?     14,  35,  42, 

21,  28,  49,  56,  63. 

Remark. — In  each  of  these  examples,  the  learner  will  perceive  that  he  di- 
vides a  number  into  equal  parts,  the  same  as  ia  division.  (Sect.  V,  Art  2, 
Obs.  1.)     Hence — Fractions  partake  of  the  nature  of  Division. 

If  a  unit  is  divided  into  tivo  parts,  each  part  is  called  one-half;  if 
into  th7'ee  parts,  each  part  is  called  one-third,  &c.     Hence — 

What  is  a  Fraction?  If  I  speak  of  an  apple,  what  do  I  evidently  rftean?  If 
I  cut  the  apple  into  two  equal  parts,  what  is  one  part  caUed  ?  If  I 
cut  it  into  three  equal  parts,  what  is  one  part  called  ?  If  I  cut  it  into 
four  equal  parts,  what  is  one  part  called?  What  are  two  parts  of  the  latter 
division  called?  Three  parts?  What  are  the  expressions  one-half,  one-third, 
one-fourth,  &c  ,  called?  How  do  we  take  one-half  of  a  number?  One-third 
of  a  number?  One-fourth  of  a  number?  In  each  of  the  given  examples  what 
doea  the  learner  do?  To  which  of  the  Fundamental  Rules  are  fractions  similar? 
If  a  unit  is  divided  into  two  parts,  what  is  each  part  called?  If  it  is  divided 
into  three  parts,  what  is  each  part  called?  From  what  then  does  the  fraction 
take  it»  JL>^m«? 

5a 


^2  COMMON    ARITHMETIC.  ScCt.  Vlll. 

Obs.  2.     Thefradlon  takes  its  name  from  the  number  of  parts  into    \ 
which  the  unit  is  divided.     Thus,  if  a  unit  is  divided  into  two  parts,     I 
each  part  is  called  one-half;  if  into   three  parts,  each  part  is  called 
one-third,  &c.     Therefore,  in  a  unit,  or  whole  number,  there  are  two 
halves,  three  thirds,  four  fourths,  nine  ninths,  twenty  tvrentieths,  &c. 

7.  How  many  fifths  in  a  unit  or  whole  number?  How  many 
sixths?     Tenths?     Fifteenths?     Twenty -ninths?     Fifty-fifths. 

Remark.  —  Every  fraction  must  have  some  value,  and  tiiis  value  also 
depends  upon  the  number  of  parts  into  which  the  unit  is  divided.  Thus  if  a  unit 
is  divided  into  3  parts,  or  thirds,  the  parts  will  evidently  be  less  than  if  divided 
into  2  parts,  or  halves;  and  if  divided  into  4  parts,  will  be  less  than  if  divided 
into  only  3.     Henee  : 

Obs.  3.  The  greater  the  number  of  parts  into  which  a  unit  is  divided 
the  less  will  be  the  value  of  each  part. 

8.  Which  is  of  the  greatest  value,  one-half  or  one-third?  one- 
fourth  or  one-fifth?     2  fourths  or  2  fiftlis?     3  fifths  or  3  sixths? 

9.  Which  is  of  the  greatest  value,  one-sixth  or  one-seventh?  one- 
tenth  or  one-twelfth?  3  eighths  or  3  ninths?  2  sevenths  or  2  eighths? 

MENTAL  EXERCISES. 

1.  What  is  one-half  of  the  following  numbers?  10,  18,  26,  40, 
84,   100,   160. 

2.  What  part  of  2  is  1?  Ans.  one-half. 

3.  What  part  of  3  is  I?  of  4?  8?  5?  7?  9?  6?  10?  12? 
20?  25? 

4.  What  part  of  3  is  2?  Ans.  two-thirds. 
Solutio7i. — One  is  one-third  of  3,  and  2  is  twice  1 ;  therefore  2  is 

twice  one-third  of  3,  and  twice  one-third  of  3  is  two-thirds  of  3. 

5.  What  part  of  6  is  2?  of  6?  7?"  8?   12?  9?  10?   14?   18?  25? 

6.  What  part  of  12  is  3?  is  4?  7?  8?  6?  5?  9?  11?   10?   15?   12? 

7.  If  half  a  bushel  of  corn  cost  2  shillings  what  would  1  bush- 
el cost?  Ans.  4  shillings. 

Solution. — There  are  two  halves  in  a  whole  number  ;  therefore, 
a  bushel  would  cost  twice  as  much  as  half  a  bushel,  and  twice  2 
are  4. 

8.  If  one-third  of  a  pound  of  raisins  costs  8  cents,  what  cost  two- 
thirds  of  a  pound?     What  cost  a  whole  pound? 

9.  If  one-fourth  of  a  pound  of  spice  cost  4  cents;  what  cost  two- 
fourths  of  a  pound?     3  fourths?     A  whole  pound? 

10.  If  1  fifth  of  a  yard  of  cloth  cost  5  shillings  what  cost  2  fifths 
of  a  yard?  3  fifths?  4  fifths?  a  whole  yard? 

What  must  every  fraction  have?  Upon  vvhat  does  this  value  depend?  If 
one  apple  were  divided  into  two  parts,  or  halves,  and  another  apple  were  divided 
into  three  parts  or  thirds,  in  Vv'hich  would  the  parts  be  the  least?  If  one  were 
divided  into  three  parts,  and  the  other  into  four  part?,  in  which  would  the  Vfl- 
ue  of  the  parts  bo  the  ieusl?     What  do  we  conclude  from  this?    r 


Art.    1.  FRACTIOKS.  83 

11.  If  1  sixth  of  a  barrel  of  older  cost  3  shillings,  ir hat  cost  2 
sixths?  3  sixths?  4  sixths?  5  sixths?  a  whole  barrel. 

12.  If  I  eighth  of  a  bolt  of  cloth  cost  one  dollar,  what  cost  2 
eighths?  4  eighths?  6  eighths?  3  eighths?  7  eighths?  6  eighths?  a 
whole   bolt?  2  bolts?  3  bolts?  4  bolts?  6  bolts?  9  bolts? 

13.  If  a  pound  of  coffee  cost  12  cents,  what  would  a  half  a  pound 
cost? 

Solution. — A  half  a  pound  would  evidently  cost  half  as  much  as 
a  pound,  and  one-half  of  12  cents  is  6  cents. 

14.  If  1  jard  of  cloth  cost  24  cents,  what  will  a  third  of  a  yard 
cost?  2  thirds?  3  yards? 

15.  If  1  acre  of  land  cost  20  dollars,  what  cost  1  fourth  of  an  acre? 
3  fourths?  2  fourths?  5  acres? 

16.  What  cost  1  sixth  of  a  ton  of  hay  at  12  ddlars  a.  ton?  3  sixths? 
6  sixths?  2  sixths?  4  sixths? 

17.  What  costs  3  sevenths  of  an  acre  of  land  at  14  dollars  an 
acre. 

Direction. — First  find  what  1  seventh  of  an  acre  would  cost  and 
then  multiply  this  by  3. 

18.  What  cost  5  ninths  of  a  pound  of  tea  at  72  cents  a  pound?  3 
ninths?  7  ninths?  1  ninth?  6  ninths?  2  ninths?  5  ninths?  8  ninths? 

19.  If  1  book  cost  half  a  dollar,  what  will  2  books  cost? 

Solution. — 2  books  would  cost  twice  as  much  as  one  book  ;  twice 

1  half  are  2  halves,  and  two  halves  are  a  unit  or  whole  number. 

Ans.  I  dollar. 

20.  If  1  man  eat  half  a  pie,  how  many  pies  will  2  men  eat?  3 
men?  4  men?  7  men?  12  men?  20  men? 

21.  If  a  boy  plants  1  third  of  an  acre  of  corn  in  a  day,  how 
much  will  he  plant  in  2  days?  3  days?  4  days?  5  days?  6  days?  7 
days^? 

22.  If  a  horse  eats  1  fourth  of  a  bushel  of  oats  in  a  day  how 
much  will  he  eat  in  3  days?  2  days?  4  days?  7  days?  5  days?  8 
days?  6  days. 

23.  What  cost  8  marbles  at  1  eighth  of  a  cent  a  piece?  What 
cost  10?   12?  16?  20?  24?  32?  40?  bQI  72?  88?  96? 

24.  If  6  apples  were  divided  equally  between  3  boys,  what  part  of 
them  would  each  boy  receive?  How  many  apples  would  each  re- 
ceive? 

Solution. — 1  is  one-third  of  5;  therefore,  each  boy  receives  one- 
third  of  the  apples.     One-third  of  6  is  2,  hence  each  boy  receives 

2  apples. 

25.  Four  men  are  to  receive  40  dollars,  each  receiving  an  equal 
share,  what  part  of  the  money  must  1  man  receive?  2  men?  3 
-nien?    How  many  dollars  must  1  man  rtcfeive?  2  men?  3  men? 


S4  COMMON    ARITHMETIC.  SeCt.   VIII. 

26.  If  6  bushels  of  wheat  cost  12  dollars,  what  part  of  that  does 
1  bushel  cost?  5  bushels?  3  bushels'?  2  bushels?  4  bushels?  ■How- 
many  dollars  does  1  bushel  cost?  2  bushels,?  6  bushels  3  bushels? 
4  bushels  ? 

27.  2  is  one-third  of  what  num*ber? 

Solution. — There  are  3  thirds  in  a  whole  number,  hence,  if  2  is 
one-third,  the  number  must  be  3  times  2,  or  6.  Ans.  6. 

28.  3  is  1  third  of  what  number?  1  fourth?  1  fifth?  1  sixth? 
1  seventh? 

29.  4  is  1  third  of  what  number?  1  sixth?  1  tenth?  1  eighth?  1 
twelfth? 

30.  5  is  1  fifth  of  what  number?  1  seventh?  1  tenth?  1 
eleventh? 

31.  4  is  2  thirds  of  what  number? 

Solution. — If  4  is  2  thirds,  one-half  of  4,  or  2,  is  1  third  of  the 
number.     Then  2X3  =  6.  Ans.  6. 

32.  Paid  3  dollars  for  3  fourths  of  a  yard  of  cloth ;  what  was 
that  a  yard?     3  is  3  fourths  of  what  number?  Ans.  4. 

33.  If  7  eighths  of  a  bushel  of  wheat  cost  84  cents,  what  is  that 
a  bushel?  84 is  7  eighths  of  what  number? 

34.  If  5  sixths  of  an  acre  of  land  costs  30  dollars,  how  much  is 
that  per  acre?     30  is  6  sixths  of  what  number? 

35.  9  tenths  of  a  dollar  is  90  cents;  how  many  cents  are  there  in  a 
dollar?     90  is  9  tenths  of  what  number? 

36.  7  ninths  of  a  hogshead  (wine  measure)  is  49  gallons;  how 
many  gallons  in  a  hogshead?  49  is  7  ninths  of  what  number? 

37.  6  is  2  thirds  of  what  number? 

38.  9  is  3  fourths  of  what  number? 

39.  10  is  2  fifths  of  what  number? 

40.  28  is  7  elevenths  of  what  number? 

41.  What  is  3  fourths  of  12? 

Solution. — 1  fourth  of  12  is  3;  then  3  fourths  is  3  X  3  =  9. 

Ans.  9. 

42.  What  is  7  eighths  of  24? 

43.  What  is  4  sixths  of  18? 

44.  What  is  4  tenths  of  40? 

45.  What  is  9  twelfths  of  24? 
i    46.  What  is  8  ninths  of  27? 

47.  What  is  8  sixths  of  18?  Ans.  24. 

48.  What  is  12  sevenths  of  35? 

49.  What  is  15  twelfths  of  24? 

50.  What  is  16  elevenths  of  33? 

Obs.  4.  When  a  number  is  divided  into  eqzial  parts,  as  in  the 
preceding  examples,  these  parts  are  called  Fractions. 

When  a  number  is  divided  into  equal  parts  what  are  these  parts  called? 


^ 


Art.  1.  COMMON  Fractions.  85 

Fractions  are  oi  three  kinds;  Common,  Decimal,  and  Duodecimal. 

1.     COMMON  FRACTIONS. 

Obs.  5.  Common  Fractions  are  expressed  hy  two  nmyxbers,  one 
written   above  the  other,  with  «  line  between  them.      Thus,  \,  |,   |, 

Obs.  6.  The  nuiriber  helow  the  line  is  called  the  Denominator,  be- 
cause it  shows  into  how  many  parts  a  unit  or  thhi^  is  divided. 

Obs.  7.  The  numher  above  the  line  is  called  the  Numerator,  because 
it  shows  the  number  of  parts  taken,  or  expressed  by  the  fraction. 

Thus,  in  the  fraction  |,  the  denominator  4  shows  that  the  unit  is 
divided  into  4  parts,  and  the  numerator  shows  that  3  of  these  parts 
are  expressed  by  the  fraction. 

Obs.  8.  The  numerator  ofnd  denoininator  taken  together  are  called 
the  Terms  of  the  Fraction. 

Remark  1.  The  number  belowr  the  line  is  called  the  denominator,  because 
it  gives  a  ndme  to  the  parts. 

2.  The  number  above  the  lin©  is  called  the  numerator,  because  it  number* 
the  partt  used. 

3.  The  word  Fraction  is  derived  from  the  Latin,  and  signifies  broken;  hance, 
fractious  are  often  called  broken  numbers.  Also,  a  whole  number  is  often 
called  an  Integer. 

Note. — The  learner  will  perceive  that  the  fraction  is  the  part,  and  not  the 
number  itself  when  broken.  Thus  if  I  break  an  apple,  it  is  evidently  a  broken 
apple,  but  only  the  piece  broken  qff"i8  the  fraction  «f  the  apple. 

Obs.  9.  It  will  be  perceived  that  Fractions  arise  from  Division^ 
•and  they  may  be  regarded  as  expressions  of  unexecuted  division;  the 
jQumerator  answering  to  the  dividend,  and  the  denominator  to  the  di- 
visor ;  the  value  of  the  fraction  being  the  quotierd  of  the  numerator 
divided  by  denominator. 

Thus,  in  the  fraction  J,  4  is  the  dividend,  2  the  divisor,  and  4  -f-  2 
=  2,  the  quotient ;  and  in  the  fraction  \,  1  is  the  dividend,  3  the 
divisor,  and  1—  3,  or  i-  of  1  the  quotient.  (Sect.  V.  Art.  2,  Obs. 
4,  a.) 

Of  how  many  kinds  are  fractions?  What  are  they?  How  are  common 
fractions  expressed?  What  is  the  number  below  tke  line  called?  Why? 
What  is  the  number  above  the  line  called?  Why?  In  the  fraction  |  which 
is  the  denominator?  Which  is  the  numerator?  What  are  the  numerator  and 
denominator  taken  together  called?  Give  another  reason  why  the  number 
below  the  line  is  called  the  denominator?  Give  another  reason  why  the  num- 
ber above  the  line  is  called  the  numerator?  From  what  is  the  term  fractfon 
derived?  What  does  it  signify?  What  then  are  fractions  often  called?  What 
is  a  whole  number,,  likewise  often  called?  From  what  do  fractions  arise? 
What  may  they  be  regarded?  To  what  do  the  numerator  and  denominator 
answer  in  division?  What  is  the  value  of  the  fraction?  In  the  fraction  1 
which  js  the  divisor,  which  thcdividend,  and  what  the  quotient?  In  the  frac- 
tion I  <j^hich  is  the  divisor,  which  the  dividend  and  what  the  quotient? 


80  COMMON  ARimMBnc.  Sect.  VJII. 

Obs.  10.  GoyiyLoyi  F&kCTiONS  3,re  either  Proper,  Imjprojy.er,  Sim- 
ple, Cojnjjound,  or   Comp)lex. 

Obs.  11.  A  Proper  Fraction  is  one  in  which  the  numerator  is 
less  than  the  denominator;  as  i->  ^,  j,  §>  y,  <fec. 

Obs.  12.  An  Improper  [fraction  is  one  in  which  the  numerator 
is  equal  to,  or  greater  than  the  denominator  ;  as  ~,  f,   Y ,  &c. 

Rkmark. — When  a  whole  number  and  h  fraction  are  expressed  togethor,  it 
i8  caliea  a  MIXED  number;  ns  2~,   S\,    5f-,    Sjj,  &C. 

Obs.  13.  A  Simple  Fraction  is  a  single  ejcpression ;  as  \,  4> 
&c.     It  may  he  either  proper  or  improper. 

Obs.  14.  A  Compound  Fraction  is  a  fraction  of  a  fraction; 
it  consists  of  several  fractions  connected  together  by  the  word  of  ;  as  |- 
of  I,  J  of  yV  of  fV,  &c. 

Obs.   15.     A  Complex  Fraction  is  07ie  which  has  a  fraction  in 

4      2\     3| 
its  numerator,  or  denominator,  or  in  both  ;  as  — ,    — ,    — ,  <Sz;c. 

5\      3      6i 

Note. — Bssidti  these  wc  hare  continued  fractions,  but  as  they  are  iiot 
QMd  iH  common  business,  they  are  not  treated  of  in  this  work. 

From  the  preceding  remarks  and  definitions,  the  demonstration  ot 
the  following  rules  is  evident: 

Obs.  16.  To  read  fractions — -First  read  the  nurnJ)er  of  parts  used 
as  shown  by  the  numerator,  and  then  the  size  of  the  parts  as  shown  by  the 
denominator. 


1.  Read  the  following  expressions: — \',  ~;  |;  j,  J;  ~\  \\;  y, 


u » 


^7.6.       8.5.       3.*»7.       7.17.11.25.19.       8.2l.2  2.yi3  6.5[> 

^8'   T'    To"'    »>    To"'    "^y"'    TT'    T2'    T3"'    T¥'    2'o"'    TI'    TT»    TTT'    ^4  aT  »    Ji 

nf      ^  ^       nf      1   l  6    .     1  fi  4  8  9  2  .     Aa    8  4  1  3    .     QQ    4  5  5  I 
'■^I     Too      OI     3040"'      ■^046"43"'     ^^T026"T'     -^^TOO-OT' 

2    \\    3|    4\        7J 

2.  Read  the  following  expressions:     — ;  — ;  — ;  — ;  . 

l\   3    i\.6\    w\\ 

3.  A  teacher  wishing  to  divide  some  fruit  among  his  scholars, sepa- 
rated it  into  a  certain  number  of  equal  parts,  and  gave  some  a  great- 
er, and  some  a  less  number  of  these  parts,  according  to  their  de- 
portment m  school.  What  John  received  could  be  expressed  by  the 
fraction  ^\ .  Now  can  you  tell  me  into  how  many  parts  the  teacher 
divided  the  fruit,  and  how  many  of  these  parts  John  obtained? 

4.  What  Charles  received  could  be  expressed  by  the  fraction  y\; 
how  many  parts  did  he  get? 

Uo-v  nro  common  fractions  divided?  What  is  a  proper  fraction?  Give 
px  i  n.»i  :v.  vvii  jt  i;^  an  improper  fraction?  Give  examples.  What  is  a  mixed 
namb^r?  (J^v'*)  r-xtJH;)!  ■•!.  What  is  a  simple  fraction?  Give  examples.  May 
it  be  proper  or  imi>i-op.=^r'i  Wbjt  ii  a  compound  fraction?  Give  examples. 
What  is  a  complex  fraciion?    Give  ex  imxlog.     How  do  we  read  fractions? 


Art.    2.  FONDAMENTAL    PROPOSITIONS.  87 

5.  What  Henry  received  could  be  expressed  by  the  friiction  #^  ; 
kow  many  parts  did  he  get? 

Obs.  17.  To  write  fractions — First  ivrite  the  numher  of  imrts 
used,  as  tJie  numerator,  and  then  the  size  of  the  parts  as  the  denomina- 
tor. 

6.  Write  two-thirds.  One-fifth.  Two-fifths.  Three-fifths.— 
Four-fifths.  Two-sixths.  Four-sixths.  One-sixth.  Three-sixths. 
Five-sixths.  Three-sevenths.  Five-eighths.  Six-ninths.  Three- 
tenths.  Eight-tenths.  Four-elevenths.  Seven-elevenths.  Nine- 
twelfths. 

7.  Write  eleven-twelfths.  Seven-twelfths.  Eleven-sixteenths. 
Eight-twentieths.  Sixteen-twentieths.  Nineteen-twenty-fourths. — 
Forty-eight-seventieths.  Fifty  units,  and  thirty-six-eighty -ninths. 
One  hundred  units,  and  four  hundred-one-thousandths.  Eighty 
units,  and  three-ten-thousandths. 

Art.  2.     Fundamental  Propositions. 

In  the  following  remarks  and  observations,  the  learner  will 
remember  that  the  numerator  answers  to  the  dividend,  the  de- 
nominator to  the  divisor,  and  the  value  of  the  fraction  to  the 
quotient  of  the  numerator  divided  by  the  denominator.  (Art. 
1,  Obs.  9.)  Thus  the  value  of  |  is  1  ;  of  |-  is  2  ;  of  J:  is  one-fourth 
of  1,  (Kc.      Therefore, 

Obs.  1.  If  the  denominator  remains  the  same,  to  mtdtiply  the  nu- 
merator by  any  numher  multiplies  the  value  of  the  fraction  by  that  num- 
her. 

Take  the  fraction  ~  =  2 ;  multiplying  the  numerator  by  f ,  w 
obtain  J  =  4,  which  is  the  same  as  2  X  2. 

Remark. — By  multiplyiog  the  dividend,  we  multiply  the  quotient.  (S«ct. 
VI,  Art.  1,  Obs.  27.  a.) 

Obs.  2.  If  the  dsnominaior  remains  the  same,  to  divide  the  numer- 
ator by  any  numher  divides  the  value  of  tlie  fraction  by  that  num- 
her. 

Take  ^  =  2;  dividing  the  numerator  by  2,  we  obtain  |=  I, 
which  is  the  same  as  2  -j-  2  =  1 . 

How  do  we  write  fraction*?  If  the  denominator  remains  the  «ame,what  f^ffcct 
does  it  have  upon  the  value  of  thf^  frnr.fion,  to  nn'li.i'y  vn  \\  \<'.\  r  lio:-  \>s  -.-mx^ 
number?  Givp  a:j  pxj  nple.  flow  Jo  w.j  kuaA'  tlii-?  prOj)osiiioii  ti  be  iru*'? 
]f  t.'"»  .'!■  i-r,  !  ijvtU.M-  rem  lius  tlie  siine  what  eifect  does  it  have  on  the  value  of 
Ji  ("n ciuri  lo  divi.JHi  the  numerator  by  any  aumber?  Givt*  ex-impios.  Why  is 
tiws  jMopcsi I ioa  correct?     What  inference  is  deduced  from  this?     Give  exani- 


88  COMMON    ARITHMETIC.  Soct.    VIII « 

Remark. — By  dividitig  the  dividend,  we  divide  the  quotient.  (Sect.  VI, 
Art.  1,  Obs.  27,  b.)     Tiierefore 

Obs.  3.  With  a  given  de')iominator,  the  greater  the  numerator,  the 
greater  the  valus  of  the  fraction.     For  |  is  greater  than  i-,  J   than  \, 

Remark. — Because,  the  greater  tho  dividend,  the  greater  the  quetient.— 
(Sect.  Vr,  Art.  1,  Obs.  26,  c.) 

Obs.  4.  If  the  numerator  remains  the  same,  to  multiply  the  denmn- 
inator  by  any  number  divides  the  value  of  the  fraction  by  that 
number. 

Take  ~=  2  ;  multiplying  the  denominator  by  2,  makes  it  J  =  1, 
which  is  the  same  as  2  -f-  2. 

Remark. — By  multiplying  the  divisor  we  divide  the  (juotient.  (Sect.  VI, 
Art.  I,  Obs.  27.) 

Obs.  5.  Jf  the  numerator  remains  the  same,  to  divide  the  denomi- 
nator by  any  number  muWqjlies  the  value  of  the  fraction  by  that  num- 
ber. 

Take  J  =  1 ;  dividing  the  denominator  by  by  2,  we  obtain  |-  =  2, 
which  is  the  same  as  1  X  2. 

Remark.  By  dividing  the  divisor  we  multiply  the  quotient.  (Sect.  VI,  Att. 
1,  Obs.  27,  a.)     Therefor© 

Obs.  6.  With  a  given  numerator  the  greater  the  denominator,  the 
less  mil  be  the  value  of  the  fraction.  For  \  is  less  than  \,  \  is  less 
than  ~,  &c. 

Remark. — Because,  the  greater  the  divisor,  the  less  the  quotient.  (Sect. 
VI,  Art.  1,  Obs.  26.) 

Obs.  7.     From  these  observations  we  notice, 

a.  1st.  It  has  the  same  ejfect  upon  the  value  of  the  fraction,  to  multiply 
the  numerator  by  any  number,  or  to  divide  tlie  denominator  by  the  same 
number. 

Take  \;  multiplying  the  numerator  by  2,  we  obtain  |  =  1 ;  divi- 
ding the  denominator  by  2,  we  obtain  y  =  1,  as  before. 

Remark. — Because,  to  multiply  the  dividend,  or  to  divide  the  divisor,  has 
the  same  effect  upon  the  quotient.     (Sect.  VI,  Art.  1,  Obs.  27,  a.) 

b.  2d.  It  has  the  same  effect  upon  the  value  of  the  fraction,  to  divide 
the  numerator  by  any  member,  as  to  multiply  the  denominator  by  the  same 
number. 

Why  is  this  correct?  If  the  numerator  remains  (he  same,  what  effect 
does  it  have  upon  the  value  of  the  fraction,  to  multiply  the  denominator  by 
any  number?  Give  an  example.  How  do  we  know  this  to  be  correct?  If 
the  numerator  remains  the  same,  what  effect  does  it  have  upon  the  value  of  the 
fraction,  to  divide  tho  denominator  by  any  number?  Give  an  example.  How 
do  we  know  this  to  be  correct?  What  inference  is  deduced  from  this?  Give 
examples.  Why  is  this  correct?  What  is  the  first  consideration  we  notice 
from  tliesc  propositions?  Give  an  example.  Why  is  this  correct?  What  is 
ihe  second  consideration  we  notice?     Give  an  example? 


Art.    2.  FUJfDAMENTAL   PROPOSITIONS.  89 

Take  ~  ;  dividing  the  numerator  by  2,  we  obtain  |  =  1  :  multi- 
plying the  denominator  by  2,  we  obtain  J  =  1 ;  as  before. 

Remark. — Because,  to  divide  the  dividend,  or  to  multiply  the  divisor,  has  the 
same  effect  upon  the  quotient.     (Sect.  VI,  Art.  1 ,  Obs.  27 , 6.) 

Obs.  8.  Multiplying  or  dividing  both  the  numerator  and  denomi- 
nator by  any  number,  changes  the  form  of  the  fraction  without  altering 
its  value. 

Take  J  =  1  ;  multiplying  both  terms  of  the  fraction  by  2,  we 
obtain  |  =  1  ;  dividing  both  terms  of  the  fraction  by  2,  we  obtain 

Rkmark. — Because  to  multiply  both  the  divisor  and  dividend  by  the  same 
number  does  not  alter  the  quotient.     (Sect.  VI,  Art.  1,  Obs.  28.) 

Obs.  9.     Jff  the  same  number  be  added  to  both  terms  of  a  proper 
fraction,  the  resulting  fraction  will  be  greater  than  the  former  fraction. 
Take  |;  adding  3  to  both  terms  of  the  fraction,  we  obtain  -f,  which 
is  greater  than  |. 

«.  ff  the  same  number  he  subtracted  from  both  terms  of  a  proper 
fraction,  the  resulting  fraction  will  be  less  than  the  former  fraction. 

Take  j,  subtracting  3  from  both  terms  of  the  fraction,  we  obtain 
\,  which  is  less  than  f . 

Remark. — These  two  propositions  are  the  converse  of  the  propositions  given 
in  Sect.  VI,  Art.  1,  Obs.  29.  When  the  terms  of  a  fraction  are  equal,  however, 
it  doesnot  affect  the  value  of  the  fraction  to  add  or  subtract  the  same  number 
from  both  terms  of  it,  because  if  equals  are  added  to,  or  subtracted  from,  equals, 
their  sums,  or  differences,  will  be  equal.     (Sect.  VI,  Art.  1,  Obs.  29,  Rem.) 

Obs.   10.  If  the  numerator  is  less  than  the  denominator  the  value  of 
the  fraction  is  less  than  1.     Thus  |-  =  one-half  of  1,  J=  three- 
fourths  of  1,  &c. 
.1    Remark. — Because  when  the  numerator  is  less  than  the  denominator,   the 
dividend  is  less  than  the  divisor. 

a.  If  the  numerator  and  denominator  are  equal,  the  value  of  the 
fractionis  1.     Thus,  |  =  1;  f  =  1  ;  |  =  1,  &c. 

Remark. — Because  when  both  terms  of  the  fraction  are  equal,  the  dividend 
and  divisor  are  equal. 
— — « ' 

Why  is  this  correct?  What  effect  does  it  have  upon  the  value  of  the  frac- 
tion to  multiply  or  divide  both  the  numerator  and  denomiator  by  the  same 
number?  Give  an  example.  Why  is  this  correct?  If  the  same  number  is 
added  to  both  terms  of  a  proper  fraction,  what  is  the  value  of  the  resulting 
fraction  compared  with  that  of  the  former  fraction?  Give  an  example?  If 
the  same  number  be  subtracted  from  both  terms  of  a  proper  fraction,  what  is 
the  value  of  the  resulting  fraction,  compared  with  that  of  the  former  fraction? 
Give  an  example.  Of  what  are  these  two  propositions  the  converse?  When 
the  terms  of  the  fraction  are  equal,  what  effect  does  it  have  upon  its  value,  to 
add  or  subtract  the  same  number  from  both  of  them?  Why?  If  the  numerator 
is  less  than  the  denominator,  what  is  the  value  of  the  fraction?  Why?  If  the 
numerator  is  greater  than  the  denominator,  what  is  the  value  of  the  fraction? 
Why?  *^If  the  numerator  and  denominator  are  equal,  what  is  the  value  of  the 
fracliftai 


00  COMMON    ARITHMETIC,  Scct.    VIII. 

h.  If  the  numerator  is  greater  than  the  denominator.,  the  value  of  the 
fraction  is  greater  than  1.     Thus   |  =  \\;  J  =  2  ;  V  =  3  ;  &c. 

RiaiARK.. — Because,  when  the  numerator  is  greater  than  the  denominator 
the  dividend  is  greater  than  the  divisor. 

Obs.  11.  Fractions  may  be  added,  subtracted,  niuliijjlied,  or  dini- 
ded,  as  well  as  whole  numbers,  but  to  perform  these  operations  it  is 
necessary  to  make  changes  in  the  form  of  the  fraction. 

Art.  3.     Reduction  of  Fractions. 

Obs.  1.  The  process  of  changing  the  terms  of  a  fraction  without 
altering  its  value,  is  called  Reduction  of  Fractions. 

Case  1 .     To  reduce  a  fraction  to  its  lowest  terms. 

Obs.  2.  A  fraction  is  reduced  to  its  loivest  terms,  when  the  numer- 
ator and  denominator  are  prime  to  each  other.  (Sect.  VII,  Art.  2, 
Note  3,  under  the  rule.) 

Ex.   1 .     Reduce  —  to  its  lowest  terms.  Operation. 

To  divide  both  terms  of  a  fraction  by  the  same     4)y|  =  |  Ans. 
number,  does  not  alter  its  value.  (Art.  2,  Obs.  8.) 

Therefore,  dividing  both  12  and  16  by  4,  we  obtain  J  as  our  an- 
swer. 

2,  Reduce  ~|  to  its  lowest  terms. 

First  Method,  Second  Methods 

6)  \\%  =  4)  II  =  2)  J-j  =  'j  Ans.     Or,  48)  HI  =  i  Anc. 

By  the  first  method  we  civide  both  192  and  240  by  6,  which  gives 
II;  next  we  divide  both  32  and  40  by  4,  which  gives  y^  >  ^^^  ^i" 
viding  both  these  by  2,  we  obtain  ^  as  our  answer. 

By  the  second  method  we  divide  both  192  and  240  by  their  great- 
est common  divisor,  (48,)  found  according  to  the  rule  in  Sect,  VII, 
Art.  2  ;  this  gives  J  for  our  answer  as  before. 

Hence — To  reduce  a  fraction  to  its  lowest  terms  : 
Obs.  3.     Divide  both  term^  of  the  fraction  by  any  number  that  will 
divide  both  of  them  without  a  remainder  ;  so  continue  to   do,  till  no 
number  greater   than   1   will  divide  bqfh  numerator  and  denotninator 
wiUiout  a  remainder.     Or, 

a.  Divide  both  terms  of  the  fractionby  their  greatest  common  divisor. 

Note  1.  When  the  terms  of  the  fraction  are  small,  it  is  generally  best  to 
work  according  to  the  first  method  ;  but  when  th«  terms  are  large^  the  latter 
method  is  tlie  most  convenient. 

Why?  Can  fractions  be  added,  subtracted,  multiplied  and  divided?  What 
is  it  necessary  to  do  in  order  to  perform  these  operations'  What  is  Reduction 
of  Fractious?  When  is  a  fraction  reduced  to  its  lowest  terms?  Wben  are 
■  umbers  prime  to  each  other?  What  is  the  irstrule  for  reducing  fractions  to 
their  iiwoat  terms?     The  soco»d  rulo?     Whon  is  it  best  to  ust  tho  Arst  rule? 


1  4 

2^T- 

Ans.  |. 

11. 

2  8  8 

Ans.  4' 

35 

Jo- 

Ans.  y^j-. 

12. 

86*4 
TTS2- 

Ans.  J. 

75 

144 

2jO-- 

Ans.  i. 

13. 

114  0 

TT2"8- 

Ans.  j%'j. 

Ans.  j%. 

14. 

4  5  3  0 

5¥3  2- 

Ans.  -^. 

4  9 
6"3 

Ans.  1. 

15. 

4  6  0  8 

JT'elT- 

Ans.  J. 

27 

Ans.  i. 

16. 

13S7  , 

1  0  0  !• 

Ans.  \~. 

44 

TT- 

Ans.  4. 

17. 

1  2  W  8 
T4-40- 

Ans.  y\. 

225 

Tooo- 

Ans.  Z^. 

18. 

43  6  a 
7409- 

Ans.  11. 

Art.  3.  RBDUCTIOxN    OF    Fxl ACTIONS.  91 

2.  Thy  learner  will  bear  in  mind  that  the  value  expressed  by  the  fraction 
does  not  depend  upon  the  magnitude  cf  its  tena«,  but  upon  the  relaliun  they 
bear  to  each  other.  Thus,  ^_'*jyL^  is  much  less  than  k,  although  its  terms  are 
larger. 

3.  The  value  of  a  fraction  is  not  altered  by  reducing  it  t«  its  iowoat  terms, 
because  the  relation  of  the  terma  to  each  other  remains  the  same.  (Art.  2, 
Obs.  8.) 

Reduce  the  following  expressions  to  tlieir  lowest  terms  : 

3. 
4. 
6. 

6. 
7» 
8. 
9. 
10. 

Case  2.     To  reduce  an  improper  fraction  to  a  whole  or  mixed  number. 

Obs.  4.  A  Mixed  Dumber  is  the  value  of  an  improper  fraction, 
expressed  hy  integers. 

Ex*  1.  Reduce  |  to  a  whole  number.  Ans.  2. 

The  value  of  a  fraction  is  the  quotient  of  the  numerator,  divided 
by  the  denominator.  (Art.  1 .  Obs.  9.)  Therefore,  6  -r-  3  = 
2.  Ans. 

2.  Express  the  value  of  V  by  integers.  Ans.  3^. 

Operation. 

In  this  case  there  was  a  remainder  after  divi-  6)21 

ding  by  6  ;  this  we  w^ite  over  our  divisor,  and  •  — 

reduce  the  fraction  to  its  lowest  terms,  as  direct-  3|  =  3|. 
ed  in  Case  1.     Hence — 

To  reduce  an  improper  fraction  to  a  whole  or  mixed  number  : 

Obs.  5.  Divide  the  numerator  by  the  denominator,  and  if  there  is  a 
remaiiuler,  write  it  over  the  divisor,  and  annex  it  to  the  quotient. 

Note. — It  is  generally  best  to  reduce  the  fraction,  (in  the  result  or  answer,) 
to  its  lowest  terms. 


Upon  what  rfoeB  the  value  expressed  liy  the  fraction  depend?  Giv«  example*. 
Is  the  value  of  th»  fraction  altered  by  reducing^  it  to  the  lowest  terms?  Why 
Mot?  What  is  a  mixed  number?  To  what  is  the  value  of  a  fraction  equal? 
What  is  the  rule  for  reducing  an  improper  fraction  to  a  whol«  or  mixed  num- 
ber? 


92  COMMON    ARITHMETIC.  Scct.  VIIL 

Reduce  the  following  expressions  to  whole  or  mixed  numbers  : 


3.  V-  ^ns.  4. 

4.  V-  Ans.  61 

5.  V"-  A^s-  4|- 

6.  V/.  Ans.  23. 

7.  Vt-  ^"s.   13yV 

8.  ^;-^  Ans.  524. 

9.   '^J^  Ans.  6U. 

10.   Vj-  ^ns.   38|. 


11.  Hr.  Ans.  30/^. 

12.  H|«.  Ans.  22||. 

14.  V\V.  Ans.  6ii?-. 

■ft  10236  Av^e:       0183 

•f/;.  4  8  321  A               in2651 

^"-  45¥T  ♦  -Ans.     lU^jg-y. 

17  8  1723  A  „jj        t1  1  1  9  4 

^'-  T5T9     •  ^nS.      Oly^y^jj-. 

18.  ^VxV'.  Ans.   18C9tVV 

J  4  5  1   1  a 


Case  3.     To  change  a  whole  or  mixed  number  to  an  improper  fraction. 

Obs.  6.  In  changing  a  ivhole  or  mixed  number  to  an  improper  frac- 
tion, its  value  must  not  be  altered. 

a.  An  integer  is  reduced  to  an  impro2:)er  fraction  by  writing  a  unit 
under  it. 

Thus,  6  =  ^  ;  10  =  V ;  19  =  T%  &c. 

Ex.  1.  Chance  6  to  a  fraction,  the  denominator  of  which  is  3. 

Ans.  y. 

Solution. — 6  =  y  ;  multiplying  both  numerator  and  denominator 
by  3,  we  obtain  y .  Ans.  Or,  there  are  3  thirds  in  a  whole  num- 
ber, or  unit,  and  therefore  in  6  whole  numbers  are  6  X  3  =  IS 
thirds,  or,  y.  Ans.     Hence — 

To  reduce  a  whole  number  to  a  fraction  with  any  given  denomi- 
nator : 

Obs.  7.  Multiply  the  ivhole  number  by  the  given  denominator  :  the 
product  will  be  the  numerator  of  the  required  fraction,  under  which 
write  the  denominator. 

2.  Chanofe  12  to  a  fraction,  whose  denominator  shall  be  9. 

Ans.  ^r- 

3.  Chanofe  15  to  a  fraction  whose  denominator  shall  be  12. 

Ans.   \%\ 

4.  Chanjre  24  to  a  fraction  whose  denominator  shall  be  36. 


Ans.   VV 


5.  Chansre  48  to  a  fraction  whose  denominator  shall  be  114. 

Ans.  VV/. 

6.  Change  |  to  an  improper  fraction. 

Solution. — 3  X  4  =  12  fourths,  and  3  fourths  make  15  fourths, 


or  y.  Ans.     Hence — 


What  must  be  ©bserved  in  changing  a  whole  or  mixed  number  to  an  impro- 
per fraction?  How  can  we  change  a  whole  number  to  an  improper  fraction? 
Give  examples.  How  do  wo  reduce  a  whole  number  to  a  fraction  with  any 
given  denominator? 


7. 

6^ 

8. 

n\. 

9. 

181. 

10. 

23  ^ 

11. 

311-. 

12. 

451. 

13. 

78tV 

14. 

99H. 

Case  4. 

Obs.  9. 

Ans 

2  5 

Ans 

85 

Ans. 

Ans. 

119 

Ans. 

1  !?1 

Ans. 

sir 

Ans. 

8  62 
TT  • 

ins.  ^ 

19*9 

r2   • 

Art.  3.  REDUCTION    OF    FRACTIONS.  93 

To  change  a  mixed  number  to  an  improper  fraction  : 

Obs.  8-  Multiply  the  whole  number  by  the  denominator  of  the  frac- 
tion, to  the  product  add  the  numerator,  and  ivrite  the  result  over  the 
denominator. 

Remark. —This  rule  is  directly  the  reverse  of  Obs.  5,  and  each  proves  the 
other. 

Reduce  the  following  expressions  to  improper  fractions  : 

15.  45^.  Ans.   V/ 

16.  8514,  Ans.  »if^ 

17.  116|^.  Ans.  5J80 

18.  240|i-.  Ans.  »W' 

19.  242i|t  Ans.  m^' 

20.  356fJ|.  Ans.   ^Vtf° 

21.  482iJ}.  ^iis-  ^  tVt' 

To  cliange  a  compound  fraction  to  a  simple  one  : 

This  process  consists  in  finding  a  single  expression  equal 
in  value  to  several  expressions. 

Ex.  1.  Change  |  of  y  to  a  simple  fraction.  Ans.  4* 

Solution.—^  of  y  is  twice  as  much  as  j  of  -y.     ^  of  y  is  /y; 

(Art.  2.  Obs.  4.)  and  |  being  twice  as  much,  is  W;  (Art.  2,  Obs. 

1.)  \\  =  ^.  Ans. 
It  will  be  perceived  that  the  result  is  obtained  by  multiplying  the 

numerators  together,  and  the  denominators-      Thus,  |  X  y  =  2 1 

The  same  process  is  observed  when  the  compound  fraction  con- 
sists of  more  than  two  terms.     Hence — 

To  change  a  compound  fraction  to  a  simple  one : 

Obs.  10.  Multiply  together  all  the  numerators  for  a  new  numerator  j 
and  all  the  denominators  for  a  new  denominaior. 

Compound  fractions  may  be  reduced  to  simple  ones  by  cancela- 
tion, which  is  often  much  shorter,  as  it  at  once  reduces  the  fraction 
to  its  lowest  terms. 

2.   Change  ^  of  f  of  y|  of  ^  of  y  to  a  single  fraction. 

Ans.  |. 


How  do  we  changt  a  mixed  number  to  an  improper  fraction?  Of  what  is 
this  rule  the  reverse?  What  relation  do  they  bear  to  each  other?  In  what 
does  the  pj-ocess  of  changing  a  compound  fraction  to  a  simple  one  consist? 
How  is  the  result  obtained  in  the  solution  of  the  first  example?        _  ,j  ,<.  :>v7H?% 


94  COMMON    ARITIIMETIC.  Scct,  VIII. 

02)eraiio7i. 

4  Since  the  r.rLnierator  of  a  fraction  answers  to  the 

5  dividend,  and  tlie  denominator  to  the  divisor,  (Art.  1. 
X$. —  /I  Obs.  9.,)  we  place  the  numerator  at  tlie  right,  and. 
;?  the  denominator  at  the  left.  In  all  other  respects,  we 
(j)  proceed  according  to  the  General  Rule,  Sect.  VIII., 
-            Art.  1.     Hence — 


8 

/7 

Ans. 

To  work  fractions  by  cancelation  : 
Obs.  11.     Place  ike  numerators  at  the  rights  and  the  denominators 
at  the  left  of  the  line;  in  all  other  respects  2)roceed  as  usual  in  canceling. 
In  the  answer,  the  numerator  is  always  at  the  right,  aad  the  denominator 
at  the  left, 

NoTK. — When  whole  or  mixed  numbers  occur,  th<?y  must  be  reduced  to  im- 
proper fractions. 

Reduce  the  following  expressions  to  simple  fractions  : 

3.  >f  J  of  ^.  Ans.  |I. 

4.  I  of  J  of  \\,  Ans.  %%\, 

5.  f  of  6  times  |  of  |-  Ans.  \\  ==  lyV- 

6.  9  times  15  times  \i  of  i-|-.  Ans.  98y\. 

7.  i  of  I  of  A  of  |.  Ans.  jU- 

8.  I  of  y\  of  15  times  423.  Ans.   1316. 
9-  4  of  Vl  of  ii  of  11-  of  ii  of  24yVp-.                   Ans,  1  .Vt- 

10.  y\  of  I  of  8 J  of  /V  of  f  of  j%  of  6J.  Ans.  2|ii. 

1 1.  I  of  I  of  A  of  A  of  i  of  i  Jf  H  of  11  ,  Ans.  J^\. 

12.  yV  of  5|  of  211  of  2V4  of  71  of  9yV  of  131. 

Ans.   10761 . 

Case  5. — To  change  a  comi)lex  fraction  to  a  sim^jle  one. 

This  process  cannot  be  thoroughly  understood  before  the  learner 
has  studied  Division  of  Fractions.  It  is  explained  in  Art.  6,  Case  5. 

Case  6. — To  reduce  fractions  to  their  least  ccmmon  dencminatcr. 

Obs.  12.  Fractions  have  a  common  denominator,  when  their  de- 
vtominaiors  are  alike,  and  in  reducing  them  their  vcdues  must  not  be 
altered. 


Is  the  same  process  obsorved  wlien  Ih©  compound  fraction  consists  of  moro 
than  two  terms?  What  then  is  the  rule  for  reducing  a  compound  fraction  to  a 
simpl*  one?  By  what  other  method  can  the  operation  be  performed?  Why 
is  this  method  preferable?  IIow  do  wa  write  the  terms  cf  the  fraction  by  this 
method?  VVhy  are  they  thus  placed?  How  do  we  otherwise  proceed?  What 
then  is  the  rule?  Wht^n  wliole  or  mixed  numbers  occur,  what  must  be  done 
with  them?  When  have  fractions  a  «omnion  dciiorninator?  What  must  bo 
observed  in  finding  their  common  denominator  ? 


Art.    8.  REDUCTION    OF    FRACTIONS.  95 

Ex*  1 .  Change  ~  and  ~  to  fractions  having  a  common  denomi- 
nator. 

Solution. — If  we  multiply  both  terms  of  a  fraction  bj^  tlie  same 
number,  its  value  will  not  be  altered,  (Art.  2.  Obs.  8.)  Therefore 
if  we  multiply  both  terms  of  the  first  fraction  (i-,)  by  the  denomi- 
nator of  the  second  fraction,  (j,)  and  both  terms  of  the  second 
fraction,  (j,)  by  the  denominator  of  the  first  fraction,  (^,)  we  shall 
obtain  ^  =  |,  and  ~  =  |.  Ans. 

2.  Change  ^,  |,  J  and  |  to  fractions  having  their  least  common 
denominator. 

It  is  evident  that  each  denominator  must  divide  the  common  de- 
nominator ;  therefore,  ike  least  common  denominator  must  be  the 
least  comTiion  multiple  of  the  denominators  of  tfie  given  fractions. 

Operation, 

We  find  the  least  common  multiple  of  l^_.^__4--6 

the  denominators  of  the  given  fractions  2 

to  be  12.     Then  2X6=12. 

^of  12=6,]  r^xi  =  TV 


^'^ f  \o       ^    r  ^^^  ^^  fractions  are  ^  ?  w  T       V* 

^  or  1^  =  J,  j  _  X  3  =  T2- 

^ofl2  =  2j  ^  UX|  =  H. 

Here  we  perceive  that  each  result  is  found  by  multiplying  both 
terms  of  each  fraction  by  the  quotient  arising  from  dividing  the  least 
common  denominator,  by  the  denominator  of  this  fraction. 

The  reason  why  we  do  this,  is,  because  we  wish  to  multiply  both 
terms  of  each  fraction  by  such  a  number  as  will  produce  an  equiv- 
alent fraction,  having  the  least  common  denominator.  Now  the  de- 
nominator of  each  fraction  must  be  one  factor  of  this  denominator, 
therefore  the  other  factor  must  be  the  number  by  which  both  ternas 
of  the  fraction  are  to  be  multiphed.     Hence — 

To  reduce  fractions  to  their  least  cgmmon  denominator  : 
Obs.  1 3,     Find  the  least  common  multiple  of  the  denominators  of  the 
given  fractions  y  and  miiUiply  both  terms  of  eauch  fraction  by  the  quUient 
arising  from  dividing  the  least  common  multiple  by  the  denominator  of 
this  fraction. 

Note. — Compound  fractions  must  be  reduced  to  simple  ones  before  finding 
their  least  common  denominator. 

The  object  of  reducing  fractions  to  a  common  denominator,  is  to 
facilitate  the  addition  and  subtraction  of  fractions,  (Art.  4.);  and  as 


Explain  the  solution  of  Ex.  1,  and  show  why  it  is  correct.  What  must 
the  least  eommon  denominator  of  several  fractions  alwaj's  be?  Why  so? 
How  do  we  obtain  each  result  in  Ex.  2?  Why  do  we  proceed  in  this  manner? 
How  theU'do  we  reduce  fractions  to  their  least  common  deuoBiiiiator?  What 
iBiitt  be  doft»  with  compound  fractioni  when  they  occur? 


96  COMMON   ARITHMETIC,  Sect.    VIIL 

it  is  generally  considered  most  practical  to  find  the  least  common 
denominator,  we  have  given  but  one  rule.* 

3.  Reduce  j,  §,  and  yo  to  fractions  having  their  least  common 
denominator. 

Note. — When  the  fractions  are  small,  the  pupil  should  be  taught  to  reduce 
them  mentally,  and  not  be  allowed  the  use  of  the  slate  or  black-board  at  all. 
This  will  accustom  him  to  thinking  for  himself. 

In  the  last  example  we  at  once  discover  that  10  will  contain  5, 
but  will  not  contain  3,  or  any  factor  of  3  ;  then  10  X  3,  or  30,  is 
the  least  common  denominator  of  the  fractions, 

Now  let  the  learner  be  questioned  thus  : 

How  much  is  j  of  30? 

If  6  is  i  of  30,  how  much  is  J? 

Then  if  |  of  30  is  24,  |  =  |J. 

Proceed  in  the  same  way  with  the  other  fractions.  The  answei*s 
are  ^  _  2  o  .        i  j_  __  24 

Reduce  the  following  expressions  to  their  least  common  denomi- 
nators : 

A       1   3   „„J   5  A„„   21   36  „^f\    35 

4.  4,  Tj-,  ana  yj.  -a.ns.  j-^,  j^-,  ana  -g-j* 

5.  f ,  ^  and  -^V-  Ans.  i|,  ||-,  and  ^V- 

6.  I.  i-,  and  1.  Ans.  f  |,  f  |.  and  \%. 


7  12   9   13  23  „nrl  3_7_ 

'•  2  5»  4  0>  60"J  1)0  J  "^^'-''  100' 


TTOO  1800»  l800»  1800»  180  IF* 


Ans. 

8.  I,  VI,  ih  ^rid  i  of  A.  Ans.  ^\,  Vro,  y'^,  and  j% 

9.  I  of  i,  4  of  I,  I  of  1,  and  f  of  y\  of  |. 

A  6  0  4  8  5  7  6  0  l_l  20  2  2_0_5_ 

JXllb,      io"Tr8  0>    T0  0  8  0"»    To  0  8  0,    TfToso* 

For  the  convenience  of  the  learner,  we  will  now  present  at 
one  view  the  following 

GENERAL  RULES  FOR  THE  REDUCTION  OF  FRACTIONS. 

To  reduce  a  fraction  to  its  lowest  terms  : 

a.  Divide  hoih  terms  of  the  fraction  hy  any  number  tliat  will  di- 
vide them  without  a  remainder ;  so  continue  to  do  till  no  number 
greater  than  1  loill  divide  them.     (Obs,  3.)     Or, 

b.  Divide  both  terms  of  the  fraction  by  their  greatest  common  di- 
visor,   (Obs,  3,  a.) 

2d,  To  reduce  an  improper  fraction  to  a  whole  or  mixed  num- 
ber: 

What  is  the  object  of  reducing  fractions  to  a  common  denominator?  What 
is  the  advantage  derived  from  solving  questions  mentally?  How  do  we  re- 
duce a  fractian  to  its  lowest  terms? 

♦Fractions  may  be  reduced  to  a  common  denominator,  by  multiplying  all  their  denomina- 
tors together  for  a  new  denominator,  and  each  numerator  into  all  the  denominators  excep*^ 
its  own  for  a  new  numerator  ;  but  this  rule  will  not  always  give  the  least  common  denomi" 
nator. 


Art.    3.  REDUCTION    OF  FRACTIONS.  97 

Divide  the  numerator  by  the  denominator  ;  write  the  remaitider^  [if 
any,)  over  the  divisor^  and  annex  it  to  the  quotient.  (Obs.  E.. 

^  3d.  To  change  a  whole  or  mixed  number  to  an  improper  frac- 
tion : 

Multijyly  the  whole  number  by  the  denominator  of  the  fraction^  to  the 
product  add  the  numerator,  [if  it  is  a  mixed  number,)  and  under 
the  result  lorite  the  denominator .    (Obs.  7  and  8.) 

4th,  To  change  a  compound  fraction  to  a  simple  one  : 

«. '  Multiply  together  all  the\numeraiors  for  a  new  numereratory  and 
all  the  denominators  for  a  new  denominator  ;  then  reduce  the  fraction 
to  its  lowest  terms.  (Obs.   10.) 

Or,  by  cancelation : 

b.  Place  the  numerators  at  the  right,  and  the  denominators  at  iheleft 
of  the  line;  then  proceed  as  usual.  (Obs.  11.) 

5th.  To  reduce  fractions  to  their  least  common  denominator  : 

First,  find  the  least  common  multiple  of  the  denominators ;  then 
midtiply  both  terms  of  each  fraction  by  the  quotient,  arising  from  di- 
viding this  multiple  by  the  denominator  of  the  fraction.  (Obs.  13.) 

EXERCISES    FOR   THE    SLATE. 

1.  Reduce  Wl%  to  its  lowest  terms.  Ans.  -J. 

2.  Reduce  |-|||  to  its  lowest  terms.  ;  Ans.  yy. 

3.  Reduce  ^ttt  ^  '^^^  lowest  terms.  Ans.  f. 

4.  Change  ^— ^to  a  whole  or  mixed  number.  Ans.  9,5^: 

5.  Change  ^|J^  to  a  whole  or  mixed  number.  Ans.   182|. 

6.  Change  144  ^|-  to  an  improper  fraction.  Ans.  ^If^. 

7.  Change  246||  to  an  improper  fraction.  Ans.   ^IJ^. 

8.  Reduce  4  of  yy  of  y|  of  fi-  of  i-|  of  J|  to  a  simple  fraction. 

•  Ans.  i^li 

9.  Reduce  J  of  \\  of  |f  of  \l  of  ^-^  of  J  of  yV  to'a'limphj 
fraction.  Ans.   gV-o- 

10.  Reduce  4,  |-,  y|,  and  y|,  to  their  least  common  denominator. 

A  1_20       112       132  i    135 

Jt4.nb.      jyy,    yyy,    yyy,    dUQ    yy y.    ^ 

11.  Reduce  J,  I,  ^2,  jq,  and  \^  to  their  least  common  denomi- 

«ofo.>.  Anc       180       210      100       13  5       „„J    144 

nator.  -tvn^s.   2^40^,  2^40^,  2^4 o">  ytq^  "'""■  '2*'^' 

12.  Reduce  y  of  \j  of  yy  and  yy  of  |-|  of  -^  to  their  least  com- 
mon denominator.  Ans.  JJ  and  |-|-. 

How   do  we  change  an  improper  fraction  to  a  whole  or  mixed  numbert 
How  do  we  change  a  whole  or  mixed  number  to  an  improper  fraction?     How 
do  we  change  a  compound  fraction  to  a  simple   one?     How  by  cancelation? 
How  4^  we  reduce  fractions  to  their  least  common  denominator? 
'    6 


99  COMMON   ARITHMETIC.  Sect.  VIII. 

Aehcie  4.    Addition  and  Subteaction  of  Fractions. 

Obs.  I.  Jhractions  are  added  or  svJbtracledy  hy  adding  or  subtracting 
heir  Tvm-^eraiors. 

MENTAL   EXERCISES. 

1 .  James  had  \  of  an  apple,  John  |,  and  Wilham  |.  How  much 
had  they  all? 

\,  and  |,  and  |  =how  many  sixths  1  Ans.  -g-  =  1. 

2.  Susan  had  f-  of  an  orange,  Mary  j,  and  Jane  \.  How  much 
had  they  all? 

3.|A  man  gave  one  boy  yj  of  a  pie,  another  -/ j,  and  another  /j. 
How  much  did  he  give  them  all? 

4.  One  man  planted  /y  of  an  acre  of  land,  another  \~,  and  an- 
other ly .     How  much  did  they  all  plant? 

5.  What  is  the  sum  of  ^%,  ~\,  IJ,  /g-,  if,  i-|,  and  |-|.l 

6.  What  is  the  sum  of  yl^,  yj^,  yVo,  tVo,  tVV,  tII>  tVo,  and 
999 

T^TT' 

7.  Matthew  has  |  of  a  pie,  and  John  \.  How  much  more  has 
Matthew  than  John? 

\  from  I  leaves  how  many  fourths?  Ans.  |  =  |^. 

8.  One  boy  has  |^  of  a  dollar,  and  another  | .  How  much  has 
the  one  more  than  the  other? 

9.  Henry  has  ^  of  a  pound  of  raisins,  and  Thomas  y.  How 
much  has  Henry  more  than  Thomas? 

10.  A.  traveled  J  of  a  mile,  and  B.  traveled  §.  How  much  far- 
ther did  A.  travel  than  B.? 

11.  From  1^  take  If . 

12.  From  J|  take  If. 

Method  of  adding  and  subtracting  fractions  hamng  different  denom- 
inators, mixed  numhers,   cfec. 

Ex.  1.  John  had  ^  of  a  dollar,  and  Frank  j  of  a  dollar.  Hot7 
much  had  both?  * 

Solution. — If  we  add  \  to  |-,  the  result  will  neither  be  |  nor  |. 
But  ^  ==  I,  and  I-  =  f  ;  and  |  +  f  =  |.  Ans. 

2,  How  much  more  had  John  than  Frank  in  the  last  example? 
SotiUim.^  =  I ;  ^  =  I ;  I  _  I  =  ^.  Ans. 

Remari:. — We  perceive  in  tkese  examples,  that  the  fractions  have  to  be  re- 
duced to  a  common  denominator  before  they  can  be  either  added  or  subtracted. 

This  principle  is  the  same  as  that  laid  down  in  Sect.  VI.  Art.  1. 
Obs.  7  and  10,    Hence — 

Obs.  2.  iVb  numbers y  whether  integral  or  fractional y  can  he  added 
■>r  suhtracted.  unless  they  are  all  cf  the  seme  name  or  Und. 

Biii'iiiiir|iiiir»f  niiiinifi.     i|        ■■  i  ^n  ■.:«■■"!  — .,.      in,-...  \.,       i.    .   .... 

livwdo  w»*44.9ir  8ubtr«?pt  fwctioas?  iixpiain  the  soiutiojis  of  Jix.  1  a.jid,^. 


Art.    4.         ADDITION    AND    SUBTftACTION    OF    FRACTIONS.  99 


3.  Add  together  |  and  J.  Ans.  yj  =  l/a- 

4.  Add  together  J  and  I.  Ans.   1||. 

5.  Add  together  j,  |,  |,  and  |.  Ans.  2j^-^. 
0.  From  J  take  |.  Ans.  |-. 

7.  From  yi-  take  y.  Ans.  /o. 

8.  From  y^  take  i-J.  Ans.  yl^. 

9.  Charles  has  12|-   dollars,  and  Henry  has  lOj  dollars.     How- 
many  dollars  have  both? 

1st  Operation. 
12-8  =  ^1-'^,  By  the  first  operation 
!0|  =  \^  =  y .  we  reduce  the  ex- 
^1-3  _|_  8_6  _  i|9  _^  235  dollars.  Ans.  pressions  to  impro- 
per fractions,  and  then  proceed  as  in  Ex.  I.      This  method  is  pre- 
ferable when  the  numbers  ars  small.  ..^  ,^^  ., 

2d  Operalhn, 

By  the  second  operation,  12^ 

ve  add  the  fractions  and  in-  lOj  =  10|     J  -|-  «  =  y  =  l|. 

tegers     separately.       This  ^ 

method  is  to  be  preferred  Ans.  23-  dollars, 
when  the  numbers  are  large. 

10.  In  the  last  example,  how   much  more  had  Charles   thaa 
Henry? 

1st  O Juration.  2nd  Operation, 

12f='f\  12J. 

103.  _   4J    _  86  IQ3    _   in  i 

i|3  __  8_p  ^  y ,  3:^  2|  dollars.  Ans,  

Ans,  Si-  dollars. 
Explanation — The  same  as  abovo,  except  i<i  read  Ex.  2,  instead 
of  Ex.  ! ,  and  subtract  instead  of  add, 

11.  Theren  has  bQ\  cents,   an(^  Clarence  has  37^^  cents.      How 
much  has  Theren  more  than  Glarenee? 

Op>eration. 
We  cannot  take  J  fromy  ;  therefore  we  5Q\ 

borrow  1   (J)  from  the  6,   and  add   it  to  37^  =  37| 

it  the  y,  making  f;  then  |-  —  I  =  y.     But  

as  we  borrowed  1  we  must  return  it;  there-  Ans.  18y  cents, 

fore  I  to  carrv  to  7  makes  8,  and  8  from  16  leaves  8,  &c. 

12.  From  78^  take  39^.  Ans.  38  ^V- 

What  principle  do  we  notice  in  these  solutions'?   With  what  does  this  principle 
oincit't;?     What  inference  is  deduced  from  this?     Explain   the  operations  of 

Ex.  9  and  10?     Wiien  ie  the  first  method  to  be  prelerred?    The  second  in«tkod? 

Expl?fin  the  operation  of  Ex.  11. 


100  COMMON    ARITHMETIC.  Scct.  VIII. 

From  the  preceding  remarks  and  illustrations,  we  derive  the  fol- 
lowing 

GENERAL  RULE  FOR  THE  ADDITION  AND  SUBTRACTION  OF 
FRACTIONS. 

JReduse  the  fractions  to  their  least  common  denominator ^  ['Rqui.,) 
and  then  add  or  subtract  their  numerators.  (Obs.  1.) 

Note. — Compound  fractions  must  be  reduced  to  simple  ones  before  they  are 
added  or  subtracted. 

To  add  or  subtract  mixed  numbers  : 

Obs.  3.  Either  reduce  them  to  improper  fractions,  and  proceed  as 
above  directed.  Or,  add  or  subtract  the  integral  and  fractional  parts 
separately. 

When  there  are  but  two  fractions  to  be  added,  and  the  numera- 
tor of  each  is  \,we  miy  add  their  denominators  for  a  numerator^ 
and  multiply  them  for  a  denominator,  and  then  reduce  the  fraction 
to  its  lowest  terms. 

Thus  ■  l-U- £  +  !<>  ^5  3 

XUU&  .    5-^  JO  —  5X10"  —  To  —  T¥' 

If  we  wish  to  subtract  fractions,  when  the  numerator  of  each  is  1 , 
loe  may  subtract  their  denominators  for  a  numerator,  and  multiply  them 
for  a  denominator,  and  then  reduce  the  fraction  to  its  lowest  terms, 

TVmc  •    ^  1     10-5  5     1  ,W 

This  is  in  effect  multiplying  both  terms  of  each  fraction  by  the 
denominator  of  the  other  fraction,  and  then  subtracting  the  nume- 
rators. 

^  Thus:    (1)  +  (^V)  =  (i  X  U)  ±  (tV  Xi)  =  m)±  {j%)  =  Vi  or 
Jo  —  To-or  Jo". 

EXERCISES    FOR    THE    SLATE. 

1 .  A  man  owning  |  of  a  vessel,  afterwards  bought  4  of  the  ves- 
sel more.     How  much  did  he  then  own?     Ans.  ||  of  the  vessel. 

2.  William  had  J  of  a  dollar,  and  his  father  gave  him  I  of  a 
dollar  more.     What  part  of  a  dollar  had  he  then? 

j\ns.   4  0  =  i^Q. 

3.  He  afterwards  spent  yj  of  a  dollar.     How  much  had  he  left? 

Ans.  I J  of  a  dollar. 

What  is  the  general  rule  for  the  addition  and  subtraction  of  fractions? 
What  must  be  done  with  compound  fractions  when  they  occur?  How  do  we 
add  or  subtract  mixed  numbers?  When  there  are  but  two  fractions  to  add, 
and  the  numerator  of  each  is  1,  how  do  we  proceed?  When  we  wish  to 
subtract  fractions,  and  the  numerator  of  each  is  1,  how  may  we  proceed? 
How,  in  reality,  are  these  two  last  cases  performed? 

-1  i  ./;';.    liJ  P 


Art.    4.         ADDITION   AND    SUBTRACTION   OF    FRACTIONS.  101 

4.  A  man  planted  4^-  acres  to  corn,   1~  acres  to  potatoes,  and 
2~-  acres  to  beans.     How  many  acres  did  he  plant  in  all? 

Ans.  7|^ 

5.  How  much  more  did  he  plant  to  corn  than  to  beans? 

Ans.   1|^. 

6.  How  much  more  to  corn  than  to  beans  and  potatoes  both? 

Ans.   2T  of  an  acre. 

7.  John  has  |  of  J  of  3  dollars,  Charles  has  |  of  |-  of  5  dollars, 
and  Luke  has  i-  of  |-  of  6  dollars.     How  much  have  they  all? 


8  How  much  more  has  Charles  than  John? 

9  How  much  more  has  Luke  than  John? 


Ans.  7~T,  dollars. 


1  2 


Ans.   1~\  dollars. 


Ans.   ly^  dollars. 
10.  A  merchant  has  three  pieces  of  cloth,  one  containing  24|- 
yards,   another   16-  yards,  and  the  other  14^  yards.     How  many 
yards  in  all?  Ans.  5o~, 

21.  How^many  yards  in  the  first  piece  more  than  in  the  second? 

Ans.  7|-. 

12.  How  many  yards  in  the  first  p'ece  more  than  in  the  third? 

Ans.  91. 

13.  How  many  yards  in  the  second  and  third  pieces  together 
more  than  in  the  first?  Ans.  6  J. 

14.  A.  has  traveled  28|-  miles,  and  B.  21|-  miles.     How  many 
miles  have  they  both  traveled?  Ans.  50—. 

15.  How  much  farther  has  A.  traveled  than  B.? 

Ans.  7^-^  miles. 

16.  Thomas  has  62\  cents,  and  Henry  has  43 J  cents.     How 
many  cents  have  both?  Ans.   106^. 

17.  How  much  more  has  Thomas  than  Henry? 

Ans.   18|  cents. 

18.  A.  has  I24y\  dollars,  B.  has  119^  dollars,  and  C.  has  214| 
dollars.     How  many  dollars  have  they  all?  Ans.  459 ^V- 

19.  How  much  has  A.  more  than  B.?  Ans.  4j  dollars. 

20.  How  much  has  C.  more  than  A.?  Ans.  90^V» 

21.  How  much  have  A.  and  B.  together,  more  than  C? 

Ans.  29^1-. 
24.  A  little  girl  had  ^  of  an  orange,  and  her  mother  gave  her  — 
of  an  orange  more.     How  much  had  she  then? 

Ans.  ~  of  an  orange. 

23.  Add  together  y'^  and  y^-  Ans.  yV- 

24.  Add  together  yV  and  y^^.  Ans.  #3*3- 

25.  Add  together  yV  and  ^1,.  Ans.  y|l^. 

26.  Add  together  ^V  and  ^|^.  Ans.  ^%%%. 

27.  ^om  yV  take  y^.  Ans,  y|y,. 

28.  From  j\  take  ^i^.  Ans.  yU-  '^^ 


Sect. 

.  VIII. 

Ans. 
Aes. 

1 

Fractions. 

i02  COMxMON    ARITHMETIC. 

29.  From  ^f  take  /„ . 

30.  From  yi-o  take  ^i^. 

Article  5.     Multipmcation  of 

Obs.  1.     In  multiplying  by  d. lyroper  fmd'ion,  the  learner  will  al- 
ways find  that  the  product  is  less  than  the  multiplicand  ;  but  this  need 
not  cause  him  any  surprise,  if  he  recollects  that  the  multiplie)^  is  less 
than  unity,  or  1  /  and  therefore,  he  only  repeats  the  multijjlicand  a 
part  of  1  time.     (Sect.  YI.  Art.  1.  Obs.  19.  Rem.) 

MENTAL    EXERCISES* 

1.  If  an  apple  costs  |  of  a  cent,  how  much  will  3  apples  cost? 
Solution. — 3  apples  will  evidently  cost  3  times   as  much  as  1  ap- 
ple ;  and  3  times  ~  equals  |,  or  1  cent.  (Art.  2.  Obs.  1.) 

Ans.   1  cent. 

2.  If  1  marble  costs  ^  of  a  cent,  how  much  will  5  marbles  cost? 

Ans.  |-  of  a  cent. 

3.  If  1  pear  costs  J  of  a  cent,  how  much  will  8  pears  cost? 

Ans.  V  =  6  cents. 
3.  At  J  of  a  dollar  per  bushel,  how  much  will  6  bushels  of  corn 
cost? 

5.  At  I  of  a  dollar  per  bushel,  how  much  would  9  bushels  of  ap- 
ples cost? 

6.  At  I"  of  a  dollar  npiece,  how  much  would  9  books  cost? 

7.  At  25  cents  a  pound,  how  much  would  |  of  a  pound  of  rai- 
sins cost? 

Solution. — The  learner  will  perceive  that  ihis  question  is  just  like 
the  preceding,  except  that  the  multiplier  and  multiplicand  have 
changed  places.  Ifow  as  it  makes  no  difference  which  factor  we 
use  as  the  multiplier,  (Sect.  IV.  Art.  2.  Obs.  5.  Hem.,)  we  may 
proceed  as  in  Ex.  1,  and  say,  25  times  |  =  V  =^  1^  cents.  Ans. 

Or,  we  may  prodeed  thus  : 

If  1  pound  cost  25  cents,  |-  of  a  pound  will  cost  \  of  25  cents,  or 
5  cents,  and  |  will  cost  3  times  as  much  as  \,  and  5  X  3=  15 
cents.  Ans. 

It  would  be  a  good  plan  to  solve  such  questions  both  ways. 

8.  A  tree  60  feet  high,  had  f^  of  its  length  broken  off  by  the  wind; 
what  was  the  length  of  the  broken  piece?  Ans.  25  feet. 

9.  The  remaining  part  was  y\  of  the  length  of  the  tree.  Required 
— the  length  of  this  part. 

What  will  always  be  noticed  in  multiplying  by  a  proper  fraction?     Need 
Why  not? 


Art.    5.  MULTIPLICATION    OF   FRACTIONS.  fw 

10.  At  50  cents  a  yard,  how  much  would  ^s  of  a  yard  of  cloth 
cost? 

11.  At  96  cents  a  pound,  how  much  will  J  of  a  pound  of  tea 
cost? 

12.  There  are  320  rods  in  a  mile.  How  many  rods  in  ^^  o^  a 
mile? 

13.  One-eighth  of  a  dollar  is  122-  cents.  How  many  cents  are 
there  in  j  of  a  dollar?  i 

Solution. — There  are  evidently  7  times  as  many  cents  as  there  are 
in  i-  of  a  dollar.  Then  7  times  12  =  84  cents,  and  7  times  ~  a  cent 
=  J  =  3i-  cents,  and  84  cents  and  Z\  cents  make  87^  cents.  Or, 
121  =  Y;  7  times  V  =  'P  =  87^  cents,  as  before. 

14.  At  16|  cents  a  yard,  how  mueh  will  6  yards  of  cs^ico  co«t? 

15.  At  18j  cents  apiece,  how  much  will  8  slates  cost? 

16.  At  Q\  cents  apiece,  how  much  will  12  lead  pencils  cost? 
SoliUion. — Here  the  multiplier  and  multiplicand  have  changed 

places ;  but  according  to  Sect.  IV.  Art.  2.  Obs.  5.  Rem.,  we  can 
proceed  as  in  the  last  three  examples.  Thus,  6  X  12  =  72  ;  \  X 
12  =  V  =  3  ;  72  +  3  =  75  cents.  Ans. 

Or,  we  can  say — 

^2  pencils,  at  6  cents  apiece,  will  cost  72  cents,  and  12  pencils  at 
J  of  a  cent  apiece,  will  cost  3  cents  ;  and  72  cents  and  3  cents  are 
75  cents,  as  before. 

It  would  be  a  good  plan  to  solve  such  questions  both  ways. 

17.  How  much  would  10  pounds  of  cojGfee  cost,  at  12^  cents  per 
pound? 

18.  How  much  would  12  pounds  of  sugar  cost,  at  8|-  cents  per 
pound? 

EXERCISES    FOR    THE    SLATK 

Case  1. — To  muUij^li/  a  fraction  and  v  whole  number  together. 

1 .  If  a  man  earn  |  of  a  dollar  a  day,  how  much  can  he  earn  in 
6  days? 

Solution.— 6  times  |  =  Y  =  '^  dollars.  (Art.  2.  Obs.  1.)  Or 
6  times  |  =  f  =  5  dollars.  (Art.  2.  Obs.  5.)       Ans.  6  dollars. 

2.  How  much  wUl  f  of  a  pound  cf  raisins  cost,  at  30  cents  per 
pound? 

Solution. — They  will  cost  |  times  30  cents  ;  bui  this  is  the  same 
as  30  times  | ;  (Sect.  IV.  Art.  2.  Obs  5.  Rem.)  30  times  f  =  V 
=  20  cents.  Ans.  Or,  i  of  30  is  10,  and  |  is  twice  10,  or  20 
cents,  as  before. 

It  will  be  perceived  that  both  these  examples  can  be  worked  by 
one  rs^.    Hence — 


104  COMMON    ARITHMETIC.  Sect.  VIII. 

To  multiply  a  fraction  and  whole  number  together  : 
Obs.  2.     Multiply  the  numerator  of  the  fraction  and  the  ivhole  num- 
her  together  J  and  divide  the  product  hy  the  denominator.     Or, 

a.  Divide  the  denominator  of  the  fraction  by  the  ivhole  number,  when 
it  can  he  done  without  a  remainder,  and  divide  the  numerator  by  the 
quotient.  (Art.  2.  Obs.  7.  a.) 

Remark. — Compound  fractions  must  of  course  be  reduced  to  simple  ones 
before  the  operation  can  be  performed. 

If  we  choOse,  we  can  work  these'  sums  by  cancelation.     Thus  : 
Ex.  1.    Operation  Ex.  2.    Operation* 


0 


^ 


10 


Ans.  5  dollars.  Ans.  20  cents. 

In  both  questions,  we  write  the  fractions  according  to  Art.  3. 
Obs.  11.,  and  the  whole  numbers  according  to  the  General  Rule, 
Sect.  YII.  Art.  1.,  and  cancel  as  usual.     Hence —  i 

To  multiply  fractions  and  whole  numbers  together  by  cancel- 
ation : 

Obs.  3.  Write  the  fraction  as  directed  by  Art.  3.  Obs.  11.,  and 
the  whole  number  as  directed  by  the  General  Rule,  Sect.  YII.,  Art.  1., 
and  cancel  as  usual. 

By  this  method  compound  fractions  may  be  reduced  to  simple 
ones  by  expression  m'^.rely. 

3.  If  J  of  a  cord  of  wood  last  a  month,  how  much  will  it  take  to' 
last  18  months?  Ans.   14  cords. 

4.  If  a  bushel  of  wheat  weigh  |  of  a  hundred  weight,  how  much 
will  24  bushels  weigh?  Ans.   14|  hundred  weight. 

5.  If  1  yard  of  cloth  cost  j|  of  a  dollar,  how  much  will  60 
yards  cost?  Ans.  48|  dollars. 

6.  8  men  spent  each  of  them  J  of  a  dollar.  How  many  dollars 
did  they  all  spend?  Ans.  6. 

7.  If  I  pay  W  of  a  dollar  for  a  bushel  of  rye,  how  much  must  I 
pay  for  28  bushels?  Ans.  22  dollars. 

8.  If  a  family  eat  yo-  of  a  barrel  of  flour  in  a  month,  how  many 
barrels  will  it  take  to  last  them  a  year,  or  12  months? 

Ans.   lOj. 

9.  If  a  merchant  sells  tea  at  J  of  a  dollar  per  pound,  how  much 
will  32  pounds  cost?  Ans.  28  dollars. 

Hov.'  do  we  multiply  a  fraction  and  whole  number  together?  Explain  why 
this  r  «thod  is  correct.  By  what  other  method  can  such  questions  be  worked? 
How  do  we  proceed  according  to  this  method?  *''3 


Art    5.  MULTIPLICATION    OF    FRACTIONS.  l05 

10.  If  a  man  spends   y|  of  a  dollar  a  day,  how  much  will  he 
spend  in  48  days?  Ans.  45  dollars. 

11.  How  much  would  i  of  a  pound  of  tea  cost,  at  96  cents  per 
pound?  Ans.  64  cents, 

12.  How  much  would  |  of  -fo  of  |-  of  a  bushel  of  wheat  cost,  at 
90  cents  per  bushel?  Ans.  72  cents. 

13.  How  much  would  J  of  |-  of  a  yard  of  cloth  cost,  at  60  cents 
per  yard?  Ans.   135  cents. 

14.  How  much  would  -^  of  an  acre  of  land  cost,  at  30  dollars  per 
acre?  Ans.  24y^y  dollars. 

Case  2. — To   mulfqyly   a  whole  number   and  a  mixed  number  to- 
gether. 

1.  How  much  would   132^   acres  of  land  cost,  at  14  dollars  an 
acre? 

Operation, 

132|  We   first   multiply  by  ^   by    14, 

14  which  gives   12*"  Next  we  multiply 

- —  132  by  14,   and  adding  the  several 

12  =  14  times  y       products  together,  we  obtain  1860 

528  dolla  s  as  our  answer.     The  reason 

132  of  our  working  this  question  thus, 

is  th«  same  as   in  Ex.   13,  Mental 

Ans.  I860. dollars.  Exercises. 

2.  How  much  would  48  yards  of  calico  cost,   at  18|  cents  per 
yard? 

Operation, 
48  This    question   is   just 

18 J  like  the  preceding  one, 

48X1=36  except   that   the   multi- 

36  =  cost  at  -  cents  per  yard.        plier   and    multiplicand 
384  have     changed    places. 

48  In  the  operation,  we  first 

.  multiply    48  by    |,   ac- 

Ans.  900  cents.  cording  to  Obs.  2  or  3, 

and  then  multiply  48  by  18,   and  adding  the  several  products  to- 
gether, we  obtain  900  cents  as  our  answer.     Hence — 

To  multiply  a  whole  number  and  mixed  number  together : 

Obs.  4.     Multiply  the  integ red  and  fractional  parts  separtely,  a7id 
add  tfieir  products  together. 

How  do-we  multiply  a  whole  number  and  a  mixed  number  together?     How 

may  we  proceed  when  the  ©ixed  number  is  small? 
6a 


105  COMMON    ARITHMETIC.  Scct.    VlII 

Hv.T.i\KK.^ — If  the  mixed  »uinber  i«  small,  we  may,  rf  we  ehoosey  reduce  it 
to  au  improper  fraction,  and  proceed  according  to  Obs.  2  or  3. 

3    How  mucli  would  26|-  yards  of  cloth  cost,  at  25  cents  per 
yard?  Ans.  659|-  cents'. 

4.  How  mucli  vfiil  47|  acres  of  land  cost,  at  16  dollars  an  acre'? 

Ans.  764  dollars. 

5 .  If  a  man  travel  1  44y^j  miles  in  a  week,  how  far  can  he  travel 
in  48  weeks?  Ans.  6934|  miles. 

6.  How  much  will  37 j  2   yards  of  cloth  cost,  at  144  cents  per 
yard?  Ans.  5388  cents. 

7.  If  3  mas  read  ^SSyl  pages  in  a  month,  how  many  pages  can 
he  read  in  24  months?  Ans.  6934|. 

8.  At  73|  dollars  apiece,  how  much  would  124  horses  cost? 

Ans.  9134|  dollars- 

9.  How  much  would  432  acres  of  land  cost,,  at  3&J  dollars  per 
acre?  Ans.   15930  dollars. 

10.  How  much  would  216  pieces  of  broadcloth  cost,  at  97— 
dollars  apiece?  Ans,  21108  dollars. 

11.  How  much  would  86  pounds  of  tea  cost,  at  93|  cents  per 
pound?  Ans,  8062|  cents. 

12.  How  much  would  74  yoke  of  cattle  cost,  at  52|-  dollars  a  yoke? 

Ans.  3912|dollars. 


'4' 


Case  3,     To  multlpl't/  fractions  together. 

Ex.  1 ,  A  man  having  \  a  dc^lar,  spent  \  of  it  for  his  dinner.  How 
much  did  his  dinner  cost  him?  Ans.  \  of  a  dollar. 

Solutimi,  If  we  divide  \j  an  apple  into  two  equal  parts,  each  part 
will  evidently  be  \  of  the  apple.  Also,  if  we  divide  ~  a  dollar  into 
two  equal  parts,  each  part  will  be  J  of  the  dollar.     Or, 

If  he  gives  ^  of  all  he  has  for  his  dinner,  and  has  but  ^  a  dollar, 
his  dinner  must  cost  him  i-  of  ^  a  dollar;  but  J  of  |-  is  a  compound 
fraction  and  equals  -.     (Art.  3,  Obs.  10.) 

2.  At  I  of  a  dollar  a  yard,  how  much  would  J  of  a  yard  of  lace 

cost? 

Solution. — If  a  yard  cost  |-  of  a  dollar,  |^  of  a  yard  will  cost  \  of 
I,  which  is  f  of  a  dollar  ;  (Art.  2.  Obs,  2.);  and  |  of  a  yard  will 
cost  3times  f,  or  J  =r|  of  a  dollar.  (Art.  2.  Obs.  1.)     Or, 

If  1  yard  costs  |  of  a  dollar,  ^r  of  a  yard  will  cost  J  of  |,  or  3V 
of  a  dollar,  (Art.  2.  Obs.  4.  and  Obs.  7,  b.;)  and  f  will  cost  |-|-, 
or  f  of  a  dollar,  as  before.  Ans.  |  of  a  dollar. 


How  do  we  multiply  fractions  together?     What  i»  the  sborlest  raethod: 
Why  so? 


Art.    5.  MULTIPLICATION   OF    FRACTIONS.  107 

It  will  be  perceived  that  the  result  in  both  these  questions  is  ob- 
tained by  multiplying  the  numerators  together,  and  the  denomi- 
nators.    Hence — 

To  multiply  one  fraction  by  another  : 

Obs.  5.  MultijLily  iogetho'  the  niifneraiors  fcr  a  new  nvfnerator, 
and  ike  denominators  for  a  new  denonunator, 

NoTK. — The  learner  will  perceive  that  this  is  precisely  like  reducing  com- 
pound fractions  to  simple  ones. 

Remark. — The  shortest  method  is  by  caneelation,  (Art.  3.  Obs.  11.),  as  it 
at  once  reduces  it  to  the  lowest  terms. 

3.  At  J  of  a  dollar  a  yard,  how  much  would  j  of  a  yard  of  linen 
cost?  Ans.  If  of  a  dollar. 

4.  At  y|  of  a  dollar  a  bushel,  how  much  would  jj  of  a  bushel 
of  wheat  cost?  Ans.  ~j  of  a  dollar. 

5.  At  \j  of  a  dollar  a  pound,  how  much  would  J|  of  a  pound 
of  tea  cost?  Ans.  |  of  a  dollar. 

6.  How  much  would  ||-  of  a  gallon  of  oil  cost,  at  yj  of  a  dollar 
per  gallon?  Ans.   1  dollar. 

7.  How  much  would  ||  of  a  yard  of  satinet  cost,  at  ^j  of  a  dol- 
lar a  yard?  Ans.  |  of  a  dollar, 

8.  Multiply  f  by  ^  ot  |-.  Ans.  i. 

9.  Multiply  yV  of  U>  by  Vt  of  ||.  Ans.  /A. 

10.  Multiply  ^  of  I  of  j%,  by  I  of  I  of  °  of  |.  Ans.  |. 

Case  4.     To  multiply  mixed  numbers  tof/ether- 

1.  Multiply  14|by  12 J. 

SoMion.—Ul  =  V  '  12|  ==  V  ;  V  X  V  ^  ^^7  Ans 
Hence — 

To  mu'tiply  mixed  numbers  together  : 

Obs.  6.  Reduce  the  mixed  numbers  to  improper  fractions ^  and 
then  proceed  according  to  Obs.  5. 

2.  How  much  would  12^  yards  of  cloth  cost,  at  5-  dollars  a 
yard?  Ans.  67|  dollars. 

3.  How  much  must  be  paid  for  18y  acres  of  land,  at  31^  dollars 
per  acre?  Ans.  590|-  dollars. 

4.  How  much  must  be  paid  for  64|  yards  of  cloth,  at  6|  dollard  a 
yard?  Ans.  412  dollars. 

5.  Multiply  8J:  by  16|.  Ans.   137^. 

6.  Multiply  18j  by  24j.  Ans.  465|-. 

7.  Multiply  34^'^2  by  48^|.  Ans.   16P2|J. 

ii^W^^f  Wf  multiply  mixed  numbers  together?. 


108  COMMON     ARITHMETIC.  Scct.    VIIL 

8.  Multiply  37^  by  62l.  Ans.  2343|.  ; 

9.  Multiply  454  by  46y\-  Ans.  2117. 

10.  Multiply  50i  by  75|f .  Ans.  3807^|-|- 

For  tlie  convenience  of  the  learner  we  now  present  at  one  view 
the  following 

GENERAL  RULES  FOR  THE  MULTIPLICATION  OF  FRACTIONS 
1st.     To  multiply  fractions  and  whole  numbers  together: 
a.  Mther  multiply  the  numerator  of  the  fraction  and  the  ivhole  num- 
ber together^    and    divide  the  product    by  the   denominator,    (Obs. 

2.)     Or, 

h.  Divide  the  denominator    by  .the  ivhole  number,  when  it  can  be 

done  loithout   a  remainder,    ami    divide  the  numercdor    by  the  quo- 

tient.  (Obs.  2.  a.)     Or,  by  cancelation: 

c.    Write  the  fraction  as  directed  by  the  4th  Ride,   Art.  3,  and 

the  ivhole    number   as    directed    by  the  General  Rule,    Sect.    YIL, 

Art.  1,  and  cancel  as  usual.   (Obs.  3.) 

2nd.     To  multiply  mixed  numbers  and  whole  numbers  togetlier:^ti 

a.  Multiply  the  integral  and  fractional  i^arts  separately,  and  add 
their  products  tQTjether.  (Obs.  4.) 

3d.     To  multiply  fractions  together  : 

Multiply  the   numerators   together  for  a  new  numerator,  and  the 
denominators  for  a  new  denominator.  (Obs.  5.) 

4th.     To  multiply  mixed  numbers  together  : 

First  reduce  them  to  improper  fractions,  and  then  proceed  as  in 
Multiplication  of  Fractions.  (Obs.  C.) 

EXERCISES    FOR    THE    SLATE. 

1.  How  much  would  24  bushels  of  wheat  cast,  at  y  of  a  dollar 
per  bushel?  Ans.  30  dollars. 

2.  How  much  w^ould  36   bushels  of  corn  cost,  at  |  of  a  dollar 
per  bushel?  Ans.  24  dollars. 

3.  If  a  store  is  worth  25000  dollars,  how  much  is  |-|  of  it  worth? 

Ans.   1357 If  dollars.    'Jtf 

4.  How  much  would  60  bushels  of  wheat  cost,  at  y  of  a  dollar 
per  bushel?  Ans.  82^  dollars. 


What  is  the  first  rule  for  multiplying  fractions  and  whole  numbers  together? 
The  second?  What  is  the  rule  by  cancelation?  How  do  we  mulliply  whole 
numbers  and  mixed  numbers  together?  How  do  we  multiply  fractions  to- 
gether?   How  do  we  multi]>ly  mixed  numbers  together? 


Art.    6.      '■^'^''^'  DIVISION    OP    FRACTIONS,  109 

5.  How  mucli  would  36  yards  of  cloth  cost,  at  ^J  of  a  dollar  per 
yard?  Ans.   25|  dollars. 

6.  How  much  would  66  books  cost,  at  J  of  a  dollar  apiece? 

Ans.  57|  dollars. 

7.  How  much  would  96  bushels  of  corn  cost,  at  j  of  a  dollar 
per  bushel?  Ans.  60  dollars.     ;b 

8.  When  land  is  worth  12  dollars  an  acre,  how  much  is  y§-  of  an 
acre  worth?  Ans.  11^  dollars. 

9.  If  powder  is  75  cents  a  pound,  how  much  is  —  of  a  pound 
worth?  *Ans.  65  cents. 

10.  How  much  would  18|  yards  of  cambric  cost,  at  16  cents 
per  yard?  Ans.  300  cents. 

11.  How  much  would  12|-  cords  of  wood  cost,  at  2  dollars  per 
cord?  Ans.  25|-  dollars. 

12.  At  87 J  cents  a  yard,  how  much  must  I  pay  for  12  yards  of 
satinet?  Ans.   1050  cents. 

13.  How  much  would  40  acres  of  land  cost,  at  15j  dollars  an 
acre?  Ans.  632  dollars. 

14.  How  much  would  72  yards  of  cloth  cost,  at  4|  dollars  a 
yard?  Ans.  328"  dollars. 

15.  How  much  would  J  a  pound  of  coffee  cost,  at  i  of  a  dollar 
per  pound?  Ans.  —  of  a  dollar? 

16.  How  much  would  y~  of  a  yard  of  cloth  cost,  atf  of  a  dollar 
a  yard?  Ans.  |  of  a  dollar. 

17.  How  much  would  41 2|-  bushels  of  corn  cost,  at  ~  of  a  dol- 
lar a  bushel?  Ans.   123||-  dollars. 

18.  How  much  would  12|  bushels  of  apples  cost,  at  31^  cents  a 
bushel?  Ans.  390|-  cents. 

19.  How  much  would  25||  acres  of  land  cost,  at  16—  dallars  an 
acre?  Ans.  41 4i- dollars. 

20.  If  a  man  travel  28|  miles  in  one  day,  how  far  can  he  travel 
in  21|  days?  Ads.  614iJ  miles.     ^ 

Article  6.     Division  of  Fractions. 

Obs.  1.  In  dividing  by  a  proper  fraction,  the  learner  will  always 
find  that  the  quotkirU  is  larger  than  the  dividend ;  but  this  need  not 
cause  any  surprise,  if  he  recollects  that  the  divisor  is  less  than  unity, 
and  consequently  the  numLer  of  parts  into  which  the  number  is  di- 
vided, must  he  greater  than  the  dividend  itself  (Sect.  VI.  Art.  1. 
Obs.  26.  h.) 


Wil&t  is  always  noticed  iu  dividing  by  a  proper  fraction?     Need  this  cause 
any  surprise?    Why  not?     How  do  we  divide  a  fraction  by  a  whole  number? 


110  COMMON    ARITHMETIC.  Scct.  VlII 

MENTAL   EXERCISES. 

If  3  pounds  of  coffee  cost  °  of  a  dollar,  how  much  is  that  per 
pound? 

Solution. — 1  pound  will  evidently  cost  \  as  much  as  3  pounds, 
and  I  of  ^  is  f  (Art.  2.  Obs.  2.) 

2.  If  5  yards  of  calico  cost  y~  of  a  dollar,  how  much  is  that  per 
yard? 

3  If  8  pounds  of  lead  cost  |§  of  a  dollar,  how  much  is  that  per 
pound? 

4.  If  6  men  own  \~  of  a  vessel,  what  part  of  it  is  owned  by 
each? 

5.  If  12  lead  pencils^cost  -—  of  a  dollar,  how  much  is  that 
apiece? 

6.  At  J  of  a  dollar  a  bushel,  how  many  bushels  of  potatoes  can 
you  buy  for  2  dollars? 

Solution. — You  can  evidently  buy  as  many  bushels  as  \  is  con- 
tained in  2.  Now  there  4  fourths  in  a  unit,  or  1  ;  then  in  2  there 
are  twice  4  fourths,  or  8  fourths  ;  that  is,  2  contains  \  8  times. 

7.  At  |-  a  dollar  a  yard,  how  many  yards  of  cloth  can  I  buy  for 
6  dollars?  -^ 

8.  At  i-  of  a  cent  apiece,  how  many  marbles  can  be  bought  for 
10  cents? 

9.  At  -  of  a  dollar  a  pound,  how  many  pounds  of  coffee  can  be 
bought  for  5  dollars?  ^1 

10.  At  ^  of  a  cent  apiece,  how  many  marbles  can  be  bought  for 
J  of  a  cent?  '  Ans.  3. 

Solution. — There  can  evidently  be  as  many  marbles  bought  as  \ 
is  contained  in  |,  and  J  contains  \,  3  times. 

11.  At  I  of  a  dollar  apiece,  how  manv  books  can  be  bought  for 
J  of  a  dollar? 

12.  At  f  of  a  dollar  a  bushel,  how  many  bushels  of  potatoes  can 
be  bought  for  Y  of  a  dollar? 

EXERCISES    FOR   I  HE    SLATE. 

Case.   1 .     To  divide  a  fraction  hy  a  whole  number. 

1.  If  4  yards  of  cambric  cost  J  of  a  dollar,  how  much  is  that  a 
yard? 

Solution. — To  find  the  cost  of  1,  we  must  divide  the  cost  of  the 
quantity  by  the  quantity.  (Sect.  VI.,  Art.  1.  Obs.  24.)  Then  |  -f- 
4  =  J  ;  (Art.  2,  Obs.  2,)  or,  |-r"4  =  /«;  (Art.  2.  Obs.  4.)  /g  == 
I,  as  before. 


Art.    6.  DIVISION    OF    FRACTIONS.  Ill 

Or,  by  cancelation :  9  I  8  __  2 

We  place  our  numbers  according  to  Art.  5,  4 

Obs.  3,  and  cancel  as  usual.  -  J  - 

9|2  =  | 
Ans.  *   of  a  dollar. 

The  several  results  it  will  be  perceived  are  alike.     Hence — 

To  divide  a  fraction  by  a  whole  number:  -r 

Obs.  2.     Divide  ike  numerator  of  the  fraction  by  the  whole  number, 

when  it  can  be  dove  without  a  remainder;  and  under  the  quotient  write 

the  denominator .     Or, 

a.  Multiply  the  denominator  of  the  fraction  by  the  whole  nuniher^ 
and  over  the  product  write  the  numerator. 

b.  Or,  by  Cancelation:     Proceed  acor ding  to  Art.  b,  Obs.  3. 

Note. — Compound  fractions,  both  in  this,  and  the  following  cases,  must  of 
course,  be  reduced  to  simple  one*  before  the  operation  can  be  performed. — 
When  the  operation  is  performed  by  cancelation,  however,  we  may  reduce 
them  by  expression,  merely. 

2.  A  man  divided  J  of  a  dollar  between  his  two  children;  how 
much  did  each  receive?  Ans.  |  of  a  dollar. 

3.  If  5  boys  can  earn  yjof  a  dollar,  how  much  can  1  boy  earn? 

Ans.  yy  of  a  dollar. 

4.  If  6  pounds  of  coffee  cost  J  of  a  dollar,  how  much  is  that  a 
pound!  Ans.  j  of  a  dollar. 

5.  If  4  bushels  of  corn  cost  Y  of  a  dollar,  how  much  is  that  per 
bushel?  Ans.  y^  of  a  dollar. 

6.  If  7  yards  of  cloth  cost  yj  of  a  dollar  how  much  is  that  per 
yard?  Ans.  ^  of  a  dollar. 

7.  If  12  yards  of  ribbon  cost  J  of  y J  of  y^  of  10  dollars,  how 
much  is  that  a  yard?  Ans.  ^  of  a  dollar. 

8.  If  15  bushels  of  wheat  cost  —^  of  100  dollars,  how  much  is 
that  per  bushel?  Ans.  J  of 'a  dollar. 

9.  If  16  pounds  of  nails  cost  J  of  a  dollar,  how  much  is  that  per 
pound?  Ans.   ^V  of  a  dollar. 

10.  If  18  lemons  cost  y^  of  a  dollar,  how  much  is  that  apiece. 

Ans.  —  of  a  dollar. 


Case  2.     To  divide  a  lohole  number  by  a  fraction. 

1.  A  teacher  divided  8   apples  among  a  cla.ss  at  school,  giving 

Why  do  we  divide  the  fraction  by  the  whole  number  in  Ex.  1,  Case  1? — 
How  do  we  divide  a  fraction  by  a  whole  number?  What  must  be  done  with 
compound  fractions  when  they  occur?     If  the  operation  is  performed  by  can- 

ceiofion  how  may  we  proceed?  wiin-.  ':    .ri.  .99!' i^^ii  • 


112  cOMMOiN  AaiTHMJ^nc.        -•        Sect.  VIII/- 

each  scholar  J  of  an  apple.     How  many  scholars  were  there  in  the 
class?  Ans.   12. 

Solution. — It  is  evident  that  there  were  as  many  scholars  as  |  is 
contained  in  8,  because  each  scholar  receives  f  of  an  apple,  and 
there  are  8  apples  to  be  divided.  We  will  first  divide  8  by  |-.  In  a 
unit  are  3  thirds,  therefore  in  8  units  are  3X8  =  21  thirds  ;  that 
is,  8  contains  ~  24  times.  Now  f  is  twice  as  much  as  ~,  consequent- 
ly 8  will  not  contain  |  but  half  as  many  times  as  it  will  ~  ;  that  is, 
12  times.  (Sect.  VI,  Art.  1,  Obs.  26,  a.)  It  will  be  perceived  in  this 
example,  that  we  multiply  the  whole  number  by  the  denominator  of 
the  fraction,  and  divide  the  product  by  the  numerator.  It  makes  no 
difference,  however,  whether  we  perform  the  multiplication  or  divis- 
sion  first;  because  8  X  3  =  24  ;  24-^-2  =  12,  and  8 -r- 2  =4;  4  X 
3=12  as  before.     Hence — 

To  divide  a  whole  number  by  a  fraction  : 

Obs.  3.     Multiply  the  xvhole  number  by  the  denominator  of  the  frac-^ 
iion,  and  divide  the  product  by  the  7iumerator.  ,i 

Or,  when  it  can  be  done  without  a  remainder, 

a.  Divide  the  whole  number  by  the  numerator,  and  multiply  the  quo- 
tient by  the  demominator. 

Obs.  4.     The  reciprocal  of  a  nundjer  in  the  quotient  arising  from 
dividing  a  unit  by  that  number.     Thus  the  reciprocal  of  2  is  \  ;  of  3, 
-',  of  5,  I,  &c.     If  we  divide  a  unit  by  ~,  the  quotient  is  y  or  2;  if. 
we  divide  a  unit  by  J,  the  quotient  is  J  =  1^.     Hence —  • 

Obs.  5.     The  reciprocal  of  a  fraction  is  the  same  fraction  inverted. 

Invert,  means  to  turn  upside  down.  Thus,  |  inverted  becomes  -J, 
&c. 

It  will  appear  from  these  definitions,  that  the  solution  of  the 
above  example  consists  merely  in  multiplying  the  whole  number  and 
the  reciprocal  of  the  fraction  tog  ether  ^  according  to  Art.  5.  General 
Rules,  Ist.'"  '"^   -   •'"• 

Examples  of  this  kind  can  also  be  performed  by  Cancelation. — 
Thus  : 

$-.4         We  proceed  as  directed.  Art.  5,  Obs.  3,  except  to 
3  use  the  reciprocal  of  the  fraction,  instead  of  the   frac- 

—         tion  itself.     Hence — 
12 

Explain  the  solution  of  Ex.  1,  Case  2.  Does  it  make  any  difference  wheth- 
er we  perform  the  multiplication  or  the  division  first?  Why  not?  How  do  we 
divide  a  whole  number  by  a  fraction?  What  is  the  reciprocal  of  a  number? 
Give  examples.  What  is  the  reciprocal  of  a  fraction?  Show  w^hy  this  is  cor- 
rect. What  does  invert  mean?  In  what  does  the  solution  of  Ex.  1,  Case  2 
consist? 


Art.  6.to»??  DIVISION  OF  fractions.-  I'l^ 

To  divide  a  wliole  number  by  a  fraction,  by  cancelation  : 
Obs.  6.     Place  the  numerator  of  the  divisor  at  the  left,  and  the  de- 
nominator at  the  right;  in  other  respects  jproceed  as  usual. 

2 .  At  J  of  a  dollar  a  bushel,  how  much  corn  can  be  bought  for 
3  dollars?  Ans.   12  bushels.     >■ 

3.  At  |-  a  dollar  a  bushel,  how  much  rye  can  be  bought  for  7  dol- 
lars? Ans.   14  bushels. 

4.  At  J  of  a  dollar  a  yard,  how  many  yards  of  muslin  can  be 
bought  for  6  dollars?  Ans.  48  yards. 

5.  At  f  of  a  dollar  apiece,  how  many  books  can  be  bought  for  8 
dollars?  Ans.   12. 

6.  At  y  of  a  dollar  a  bushel,  how  much  barley  can  be  bought  for 
12  dollars?  Ans.   16  J  bushels. 

7.  A  man  gave  12  dollars  to  some  destitute  people,  giving  each 
Y  of  a  dollar.     How  many  were  relieved?  Ans.   14. 

8.  At  i-  of  I  of  10  dollars  a  bushel,  how  much  wheat  can  be 
bought  for  18  dollars?  Ans.   14|  bushels. 

In  this  example,  the  expression  J  of  |  of  10  dollars  is  our 
divisor. 

9.  At  y  of  a  dollar  apiece,  how  many  hats  may  be  bought  for  6 
dollars?  Ans.  21. 

10.  At  —  of  J  of  J  of  12  dollars  apiece,  how  many  slates  can  be 
bought  for  9  dollars?  Ans.  36. 

Case  3.     To  divide  one  fraction  hy  another. 

1.  At  f-  of  a  dollar  a  yard,  how  many  yards  of  cloth  can  be 
bought  for  I  of  a  dollar?  Ans.   \~. 

The  result  is  evidently  found  by  dividing  |  by  |. 

Operation. 

f  =  |.         We  first  reduce  the  fractions  to  a  common  denomina- 
|-  -T- 1  =  1^.   tor,  and  then  divide  the  numerator  of  the  dividend, 
by  the  numerator  of  the  divisor. 

We  can  however,  divide  one  fraction  by  another  without  reducing 
them  to  a  common  denominator. 

We  learn  from  the  remark  under  Obs.  6,  that  to  divide  a  whole 
number  by  a  fraction,  consists  merely  in  multiplying  the  whole  num- 
ber by  the  reciprocal  of  this  fraction.  Hence,  it  is  evident  that  to 
divide  one  fraction  by  another  consists  merely  in  multiplying  thefrac- 

How  do  we  divide  a  whole  number  by  a  fraction,  by  cancelation?  Explain 
the  solution  of  Ex.  l,Case  3.  Is  there  any  other  method  of  performing  such 
operations?     In  what  do  such  operations  consist? 


114  IbOMMON    ARITHMETI0.  Scct.    VIII. 

Hon  which%is  the  dividend, dy  (ha  reciprocal  »/  the  fraction  which  is  the 
divisor.  That  is,  we  invert  our  divisor  and  proceed  according  to  Art. 
3,  Obs.  5,  thus:  ;-  X  I  =  ;4  =  I  =  li-.  Ans. 

Or,  we  may  proceed  hj  oancelation,  according  to  Art*  3,  Obs.  11, 
thus: 

3__^  ^._4  ITie  learner  will  perceive  that  we   write  our 

^  ^  divisor  as  directed  by  Obs.  6.     In  all  these  op- 

—  —  erations  the  final  result  m  the  same.     Hence — 

3  4=:lJ.Ans. 

To 'divide  one  fraction  by  another  : 

Obs.  7.     Invert  ike  divisor  and  then  proceed  as  in  nwltiplication  qf 
fractions.     Or,  by  cancelation: 

a.    Write  the  divisor  as  directed  bi/   Obs.  6,  and  the  other  mcnlws 
as  directed  by  Art.  3,   Obs.  1 1,  and  cancel  as  usuall 

2.  If  a  man  mow  ~  of  an  acre  of  grass  per  day,  how  many  days 
will  it  take  him  to  mow  y|  of  an  acre?  Ans.  3J. 

3.  If  a  man  traf'el  |  of  a  league  in  an  hour,  how  many  hours  will 
it  take  him  to  travel  y^  of  a  leaguel  Ans.  j  of  an  hour. 

4.  At  y  of  a  dollar  a  yard,  how  many  yards  of  cloth  can  I  buy 
fory|-  of  a  dollar?  Ans.  l^j. 

5.  If  I  pay  4  of  a  dollar  a  pound,  how  many  pounds  of  tea  can 
I  buy  for  J  J  of  a  dollar?  Ans.   J  of  a  pound. 

6.  How  many  yards  of  satinet  can   I  buy  for  -J  of  a  dollar,  at 
yy  of  a  dollar  a  yard?  Ans.  |  of  a  yard. 

7.  Divide  f  of  J  of  4,  by  ?  of  4  of  J  of  7.  Ans.  y^. 

The  learner  will  observe  that  he  must  invert  ail  the  terms  of  the  di- 
visor, when  it  is  a  compound  fraction. 

8.  Divide  ^  of  |  of  |  of  ^  of  52  by  |J  of  J  of  J  of  28, 

Aas.   1. 

9.  Divide  }  of  J  of  y\  of  i|  of  23  by  ?  of  JJ  of  -\|  of  46.     . 

Aas.  ^  -^ , 

10.  Divide  i-  of  4  of  y"^  of  20,  by  |  of  -J  of  -,\  of  32.     '^'^k 

Ans.  laads 
Case  4,     Method  of  proceeding  when  mixed  numbers  occur.      T 

1.  How  many  bushels  of  wheat  can  I  buy  for  b\  dollars,  at  1  j 
dollars  per  bushel?  Ans.  4J. 

Solution.— b\  =  V  ;  U  =  T  ;  V  -V-  ¥  =  44  bushels,  Ans. 

How  do  we  divide  one  fraction  by  another?     How  by  cancelation?     When 
our  divisor  is  a  compound  fraction,  what  must  be  observed  in  the  operation? 


Art.    G.  DIVISION   OF    FRACTIONS.  115 

Hence — When  mixed  numbers  occur  in  both  the  dividend  and 
divisor : 

Obs.  8.  Reduce  them  to  improper  fractions ,  and  then  pr<X4^cd  ojc- 
cordingto  Obs.  7. 

2.  How  much  rye  can  I  buy  for  4\j  dollars,  at  yy  of  a  dollar  per 
bushel?  Ans.  8  bushels 

3.  How  many  books  can  I  buy  for  1  \\  dollars^  at  j^  of  a  dollar 
apiece?  Ans.  2. 

4.  At  5\  dollars  an  acre,  how  many  acres  of  land  can  I  buy  for 
52^  dollars?  Ans.   10. 

-    5.  At  2y  dollars  apiece,  how  many  sheep  can  I  buy  for  S2j  dol- 
lars? Ans.   15. 

6.  At  24j  dollars  per  acre,  how  many  acres  of  land  can  I  buy  for 
620  dollars?  Ans.  25. 

Solution.— 24^  =  *!'  ;  620-f-'|'  =  25  acres,  Ans. 
Hence — ^When  only  the  divisor  is  a  mixed  number: 
Obs.  9.     Reduce  it  to  an  im2:>roper  fraction,  and  t/ien  proceed  accor- 
ding to  Obs.  3,  or  6. 

7.  At  52^  dollars  apiece,  how  many  horses  can  I  buy  for  2220 
dollars?  Ans.  42. 

8.  At  17^  dollars  apiece,  how  many  cows  can  I  buy  for  314  dol- 
lars? Ans.   18. 

.  9.  At  93 J  cents  per  bushel,  how  many  bushels  of  wheat  can  I 
Duy  for  1875  cents?  Ans.  20. 

10.  At  62l  cents  per  bushel,  how  many  bushels  of  apples  can  I 
buy  for  3000"  cents?  Ans.  48. 

11.  At  3  dollars  a  yard,  how  many  yards  of  cloth  can  I  buy  for 
187^  dollars?  "  Ans.  62i-. 

Solutio?i.—lQ7\  =  ^' ;  ^15  -^3  =  »|^  =  62^.  Or,  187^^  -^  3 
=  62,  and  1|-  remainder  ;  li-=|;  ^-^-3=|■,  which  annexed  to 
62  =  62|.     Hence— 

When  only  the  dividend  is  a  mixed  number : 

Obs  10.  Reduce  it  to  an  Improper  fraction,  and  then  p)^'Oceed  accor- 
ding to  Obs.  2.     Or, 

a.  Divide  the  integral  part  as  in  simple  division;  to  the  remainder 
{if  any)  annex  the  fractional  part  and  proceed  as  above  directed, 

Note. — The  first  rule  is  the  best  when  the  dividend  is  small,  but  when  the 
dividendis  large,  the  latter  method  is  preferable. 

When  both  dividend  and  divisor  are  mixed  numbers,  how  do  we  proceed? — 
When  only  the  divisor  is  a  mixed  number,  how  do  we  proceed?  When  only 
the  dividend  is  a  mixed  number,  how  do  we  proceed?  When  is  the  first  rule 
prefcBable  in  this  case?  The  second?  What  must  be  ©bserved  in  all  cases  of 
redu'ctloa  of  fractions? 


iia 


COMMON  ARITHMETIC. 


Sect.  VIII. 


12.  How  many  acres  of  land  can  I  buy  for  146|  dollars,  at  8  dol- 
lars an  acre?  Ans.  1S~. 

13.  How  many  lead  pencils  can  I  buy  for  37^-  cents,  at  3  cents 


apiece 


Ans.   n)y. 


14.  At  15  dollars  an  acre,  liowmany  acres  of  land  can  I  buy  for 
1234i  dollars?  Ans.  S2^%. 

1 5.  At  45  cents  a  yard,  how  many  yards  of  cloth  can  be  bought 
for  843 J  cents?  Ans.  18  J, 

Case  5.      Complex  Fractions. 

Note. — This  and  the  following  case  properly  belong  to  Reduction  of  Frac- 
tions, but  it  was  thought  best  to  defer  them  until  the  learner  had  studied  Di- 
vision of  Fractions.  .^, 

Remark. — The  learner  will  recollect,  that  in  all  cases  of  reduction  of  frac- 
ions,  the  value  of  thefraciion  must  not  be  altered.     (Art.  3,  Obs.  1.)  ■' 


2^- 
^1.  Reduce   —  to  a  simple  fraction. 
4^ 


Ans.  Y^, 


Solution. — This  expression  is  the   same  as  2^- -i- 4j.     (Art,  1. 
Obs.  9.)     By  Obs.  8,  we  find  the  quotient  of  2^  divided  by  ^\  to  be 


i-|.     Hence— 


To  reduce  a  complex  fraction  to  a  simple  expression  :  .  .,'^[ 

Obs.  11.  Reduce  both  numercdor  and  denominator  to  improper 
fractions,  and  then  divide  the  former  expression  by  the  latter^  as  directed 
by  Obs.  7. 


Change  the  following  expressions  to  their  simplest  form: 


2.  — . 
2| 

8| 


4.  — 
6 


61 


Ans.2  f 

1 
6, 

Ans,  2|, 

7, 

Ans.  1-V. 

8. 

Ans,   ll-|. 

9, 

M 


15 


Ans.  f 


31- 


24| 


4^ 


Ans,  25. 


Ans.  2.  -^y) 


Ans.  6, 


How  do  we  reduce  a  complex  fraction  to  a  simple  expression? 


'•U^kiT 


Art.  6. 


DIVISION    OF    FRACTIONS. 


117 


4 
10.  — 


11. 


12\ 


Ans,  1^. 


Ans,  j%'j. 


6^ 
12.   5  — 


13. 


u 


Ans.  t'o 


Remark  1. — Complex  fractions  may  be  either  added,  subtracted,  multiplied 
or  divided,  by  first  reducing  them  to  simple  expressions. 


14. 

4|           61 
Add  together .  — ,  and  — . 
1^          31 

15. 

4  e\        2| 

Add  together  — ,  — ,  and  — , 
11     9            3| 

16. 

Add  together  ,  — ,  and , 

2tV    3i           711 

17. 

7A             31 
From take  . 

9A            811 

161             4 

18. 

From take . 

9             10| 

19. 

From take . 

29- 


Ans    41^^ 


Ans.  4|i 


K^„       19  467 


A „„       12385 
Ans.     3TT7  9 


Ans.  lf|. 


Ans-  yVy. 


Remark  2. — When  complex  fractions  are  multiplied  or  divided  by  cancelation, 
we  may  merely  reduce  them  by  expression,  and  proceed  according  to  Obs.  7,  or 
Art.  5,  Obs.  5. 

7o  Ov  '*  "  ' 


20.  Multiply by 

9-r\ 


Ans.  j\%. 


n 


21.  Multiply    —  by 


5i 


Ans.  21. 


Can  complex  fractions  be  added,  subtracted,  multiplied,  or  divided*?  How? 
When  complex  fractions  are  multiplied  or  divided  by  cancelation,  how  may  we 
proce/^dl? 


118  COMMON    ARITHMETIC.  Sect.    VIIj 


22.  Multiply 

4i 

.         HI 

-  by  .                                       Ans.  G^V 

6| 

21           7 

23,  Multiply    —  by  — .                                                 Ans.  |^, 

S          4| 
1-         2- 

24.  Multiply    -^  by  ~.                                                Ans.  U. 

3t         4L 

24|         41 

25.  Divide  by  —                                                      Ans.  6, 

81          81 

12 

26.  Divide  —  by 
4i 

5i 

— .                                                 Ans,  S^V. 
6 

4^ 
27.  Divide  —  by 
3| 

8^ 

-.                                                 Ans.  /j. 

It 

li 

28.  Divide  —  by 

2i 

2i 

— .                                                  Ans.  1|. 

10 

29.  Divide  —  by 
31 

4 

-.                                                 Ans.  |i. 

2| 
30.  Multiply   —  1 
5\ 

6|             .                           8tV 
Dy  and  divide  product  by .  Ans.  1. 

HA                                           401 

Case  6. — To  change  a  fraction  to  a  required  denominator. 

This  process  consists  merely  in  multiplying  both  the  terms  of  the 
given  fraction,  by  such  a  number  as  will  give  a  resulting  fraction 
having  the  required  denominator.     Hence — 

Obs.  12.  The  denominator  of  the  given  fraction,  is  a  factor  of  the 
denominator  of  the  required  fraction. 

In  this  case,  our  multiplier  (which  must  be  the  other  factor, )  is 
found  by  dividing  the  denominator  of  the  required  fraction,  by  the 

In  what  doei  the  process  of  changing  of  a  fraction  to  any  required  denomi- 
nator consist?    What  inference  is  deduced  from  this? 


Art.  6.      i  DIVISION  or  fractions.  1X9 

denominator   of  the   given  fraction.     (Sect.  VI.  Art.  1.  Obs.  16.) 

1.  Change  \  to  a^fraction,  the  denominator  of  which  is  8.  Ans.  j. 

Solution. — Dividing  8  by  4,  we  obtain  2  as  our  multiplier,  and 
multiplying  both  terms  of  the  fraction  (\)  by  2,  we  obtain  |  as  our 
answer. 


n- 

2.  Change  §■  to  twelfths.  Ans.  — 

12. 


As  the  denominator  of  the  required  fraction  is  always  given,  our 
only  trouble  is  in  finding  the  numerator.  In  the  last  example,  our 
numerator  is  found  by  dividing  12  by  5,  and  multiplying  the  quotient 
by  o.  It  is  evident  that  by  first  performing  the  multiplication,  and 
then  the  division,  we  shall  obtain  the  same  result,  and  thus  avoid  a 
fractional  multiplier.     Hence — 

To  change  a  fraction  to  any  required  denominator : 

Obs.  13.  Multiply  the  numerator  of  the  given  fraction  hy  the  de~ 
-nominator  of  the  required  fraction,  and  dimde  tht product  by  the  denom- 
inator of  the  given  fraction  ;  the  result  will  be  the  numerator  of  the  re- 
quired fraction  under  which  write  the  required  denominator. 

3.  Change  |  to  thirty -fifths.  Ans.  f  J. 

4.  Change  ^^  to  fifteenths. 

6.  Change  y  ^  to  thirds. 


Change  4  to  fifths. 


7.  Change  J  to  elevenths. 

8.  Change  I  to  fourths.  Ans. 


In  this  ca«!e  what  must  our  multiplier  be?     How  is  it  found?     In  what  does 
the  chief  ciifficulty  lie,  in  solving  questions  of  ibis  nature?     How  do  wo  find 


Ans. 

4-i 

16 

Ans. 

l^\ 

3 

^ 

5 

Ans. 

Si 

11 

2| 

the  nuibjorator  in  Ex.  2? 


■vfeaiaica'jb  i;«^i;,. 


9.  Change  yy  to  ninths. 


.COMMON    ARITHMETIC.  Ssct.    VIII. 

9 


10.  Chanofe  A  to  sevenths. 


1  5 


7 


For  the  convenience  of  the  learner,  we  will  now  present  at  one 
view  the  following 

GENERAL  RULES  FOR  THE  DIVISION  OF  FRACTIONS. 

1st.  To  divide  a  fraction  by  a  whole  number  : 

a.  Divide  the  numerator  of  the  fraction  hy  the  whole  number,  when 
it  can  he  done  without  a  remainder ,  and  under  the  quotient  write  the  de- 
nominator.    (Obs.  2.)     Or, 

h.  Multiply  the  denominator  of  the  fraction  hy  the  whole  number, 
and  over  the  product  write  the  numerator.  (Obs.  2,  a.)  or  bycancel- 
ation: 

c.  Proceed  in  every  respect  as  in  multiplication  of  fractions. — 
(Obs.  2,5.) 

2d.  To  divide  a  whole  number  by  a  fraction  : 

a.  Multiply  the  whole  number  hy  the  denominator  of  the  fraction 
and  divide  the  product  hy  the  numerator.  (Obs.  3.)  Or,  when  it 
can  be  done  without  a  remainder. 

h.  Divide  the  whole  number  hy  the  numerator,  and  multiply  the 
quotient  by  the  denominator .     (Obs.  3.  a. j     Or,  by  cancelation. 

c.  Proceed  in  every  respect  as  in  multplication  of  fractions,  except  to 
write  the  numerator  of  the  divisor  at  the  left,  and  the  denominator  at 
the  right.     (Obs.   6.) 

3d.  To  divide  one  fraction  by  another: 

Invert  the  divisor  and  then  proceed  as  in  multiplication  of  fractions. 
(Obs.  7.)     The  rule  by  cancelation  is  the  same.     (Obs.  7.) 

4th.  When  mixed  numbers  occur  either  in  the  dividend,  or  di- 
visor, or  both. 

Reduce  them  to  improper  fractions,  and  proceed  according  to  pre- 
ceding rules.     (Obs.  8,  9  and  10.) 

6th.     To  reduce  a  a  complex  fraction  to  a  simple  one: 

Consider  the  numerator  a  dividend^  and  the  denominator  a  divisor, 
and  then  proceed  according  to  the  last  rule.     (  Obs.   11.) 

What  is  the  rule  for  dividing  a  whole  number  by  a  fraction?  The  rule  by 
cancelation?  What  is  the  rule  for  dividing  one  fraction  by  another?  What  is 
the  rule  when  mixed  numbers  occur?  What  is  the  rule  for  reducing  a  com- 
plex fraction  to  a  simple  one?  What  is  the  rule  for  changing  a  fraction  to  any 
required  denominator? 


Art.    6.  PIVK^ION    OF    FRACTIONS.  l2^1i 

6 til.     To  change  a  fraction  to  any  required  denominator. 

Multiply  the  numerator  of  the  given  fraction,  hy  the  denominaior  of 
the  required  fraction,  and  divide  the  product  hy  the  denominator  of  the 
given  fraction  ;  the  result  will  be  the  numeratoi'  of  the  required  frac- 
iion,  under  tohich  write  the  denominator.     (Obs.  13.) 

EXERCISES    FOR    THE    SLATE. 

1 .  If  6  slates  cost  J  of  a  dollar,  how  much  is  that  apiece? 

Ans.  ^  of  a  dollar. 

2.  If  14  yards  of  ribbon  cost  |j  of  a  dollar,  how  much  is  that 
per  yard?  A.ns.  ~  of  a  dollar. 

3.  At  J  of  a  dollar  per  bushel,  how  many  bushel  of  com  can  I 
buy  for  10  dollars?  Ans.  40. 

4.  At  —  of  a  dollar  per  bushel,  how  many  bushel  of  oats  can 
be  bought  for  75  dollars?  Ans.  400. 

5.  How  many  yards  of  cloth  at  |  of  a  dollar  a  yard,  can  be 
bought  for  6  dollars?  ^  Ans.   16. 

6.  How  many  books  at  |  of  a  dollar  apiece,  can  be  bought  for 
10  dollars?  Ans.   12, 

7.  How  many  tumblers  at  fV  of  a  dollar  apiece,  can  be  bought 
for  15  dollars?  Ans.  48. 

8.  How  much  wheat  can  be  bought  for  15  dollars,  at  Ij  dollars 
per  bushel?  Ans.   1 2  bushels. 

9.  At  ^  of  a  dollar  apiece,  how  many  slates  can  I  buy  for  y  of 
a  dollar?  Ans.  11. 

10.  At  yg  of  a  dollar  a  roll,  how  many  rolls  of  tape  can  be 
bought  for  I- of  a  dollar?  Ans.   14. 

11.  A  t  8y  cents  a  pound,  how  many  pounds  of  coftee  can  be  bought 
for  87  I  cents?  Ans.   10|. 

12.  At  16|  cents  a  pound,  how  many  pounds  of  spice  can  be 
bought  for  93|  cents?  Ans.  5|-; 

13.  How  many  boxes  wiUit  take  to  contain  1600  pounds  of  tea, 
each  box  containing  44|  pounds?  Ans.  36. 

14.  How  many  barrels  of  pork  can  be  bought  for  478j  dollars, 
at  13y\-  dollars  per  barrel?  Ans.  36. 

15.  How  many  bales  of  velvet  in  1166|  yards,  each  bale  contain- 
ing 1201^  yards?  Ans.  9 


16.  Add  together 

1^ 
2^' 

3\          4 
—  and  — . 
5\         6? 

Ans.  iHfl 

17.  Add  together 
7 

2f 
3|' 

4i          15 
—  and  — . 

6        Mr 

Ans.  4#A, 

122  COMMON    ARITHMETIC.  Scct.    VIII. 


18. 

3^            li 
From     —  take  — . 

41           21 

15           44 

19. 

From     —  take  — 

5J             6 

20. 

71          3-? 
Multiply —  bv  — . 

4e      81 

21. 

5           11 

xMultiplv —  by  — . 
■   4|         2| 

2A        1- 

22. 

Divide  bv  — . 

5          10 

23. 

Divide  —  by  — . 

41-          1^^ 

Am-.  U- 


Ans. 


.iv 


Ans.  fl5. 

Ans.  \-i 

Ans.  21  J. 

uod 

Ans.  I.- 

or 

10| 

Ans. . 

13 

211 
Ans. . 

25 

25.  Cliano-e  1  to  twentv-fifths. 


26.  If  41  bushels  of  wheat  cost  3|  dollars,  what  will  Gj  bushels 
cost?  Ans.  6  jl  dollars. 

First  find  the  cost  of  1  bushel,  and  then  of  6~ 

27.  If  4  men  spend  17  dollars  in  5j  days,  how  many  dollars  will 
7  men  spend  in  16|  days?  Ans.  90f  J. 

Solution. — If  4  men  spend  17  dollars  in  51  days,  1  man  will 
spend  17 -—4=  4l  dollars  in  51  days  ;  and  he  will  spend  4l-^-5l 
=  IJ  of  a  dollar  per  day.  Then  7  men  will  spend  7  times  11,  or 
W  dollars  per  day,  and  VV  X  16j=  901 J  dollars  in  IGj  days. 

28.  A  man  after  speniin^r  1, 1  and  \  of  his  money,  found  he  had 
26  dollars  left;  how  much  had  he  at  first?  Ans.  120  dollars. 

Solution. — A  riding- 1^,  \  and  tt  toarether,  we  find,  their  sum  to  be 
41.     Thi|  is  whMM^e  spent.     Then  1 JI  _  ^I  n=  2.J  \.  what  he  had 


Art.    G.  DIVIS.ON    OF    FRACTIONS.  128| 

20.  If  a  certain  number  is  incj:eased  by  J,  |-,  |,  |  and  |-  of  itself, 
and  33  more,  the  result  will  be  4  times  the  original  number;  what  is 
this  number?  .  Ans.   180. 

Solution. — Addino-  together  \.  \,  \,  |,  and  ~,  we  find  their  sum 
tobe'^;  This  added  to  1,  oi" JJ  =  ^ -3^i.  4  — 34J--il; 
then  33  is  ~\  of  what  number? 

MISCELLAKEOUS     EXERCISES    EOR    THE    SLATE: 

Involving  the  princijyles   of    Cotnmon    Fractions. 

i.  At  I  of  a  dollar  a  yard,  bow  much  Vr'ould  |  of  a  yard  of  dotli 
cost?  Ans.  II  of  a  dollar. 

2.  If  J  of  a  yard  of  cloth  co:it  \~  of  a  dollar,  how  much  is  that  per 
yard?  Ans.  -g- of  a  dollar. 

3.  If  6  pounds  of  cotFec  cost  J  of  a  dollar,  bow  much  is  that  per 
pound?  Ans.  -  of  a  dollar. 

4.  If  $.\  yards  of  clotb  cost  ~  of  a  dollar  bow  much  would  Q~ 
yards  cost?  Ans.   if-^  dollars. 

5.  How  much  would  12|  acres  of  land  cost,  at  8|-  dollars  an  acre? 

Ans.   105  dollars, 

6.  If  Ya  Oi^  a  ship  cost  4728  dollars,  bow  much  is  ~  of  her 
worth?  Ans.  6910  dollars. 

7.  If  ~  of  yo  of  a  store  is  worth  2448  dollars,  what  is  the  worth 
of  the  store?  Ans.  3400  dollars. 

8.  If  7^^  bushels  of  wheat  cost  8|  dollars,  bow  much  w^ill  \Q~ 
bushels  cost?  Ans.   19^  dollars. 

9.  If  a  pole  9~  feet  bigbcast  a  shadow  12-^-  feet,  bow  high  must 
that  pole  be  that  casts  a  shadow  of  147  feet? 

Ans.   114  feet. 

10.  If  9  men  can  do  apiece  of  work  in  1C§  days,  in  wba.t  time 
can  14  men  perform  it?  „Ans.  lOy  days. 

11.  If  it  requires  3^  yards  of  clotb  to  make  a  coat,  when  it  is  but 
J  of  a  yard  wide,  bow  much  will  it  require  wdien  the  cloth  is  1  yard 
wide?     IIow  much  wbentbe clotb  is  1^  ^rds  wide? 

Ans.  to  the  last.  2yV  yards. 

12.  If  9  horses  consume  5|-  tons  of  hay  in  7  w'eeks,  how  many 
tons  will  16  horses  consume  in  12  weeks.  Ans.   17^. 

13.  If  6  students  spend  b~  dollars  in  12y  days,  bow  many  dollars 
will  15  students  spend  in  23i-  days.?  Ans.  24|. 

14.  If  a  family  of  7  persons  drink  18|-  gallons  ot  beer  in  2^-  weeks, 
bow  many  gallons  will  they  drink  in  12|  weeks,  if  7  persons  more 
are  added  to  the  family?  Ans.   187. 

1 5.  A  man  spent  \,  \,  and  \  of  bis  money,  and  bad  69  dollars  • 
left,  bow  much  had  he  at  first?  Ans.   180  dollars. 

.   16.' A  mau  has  an  orchard  in  which  \  of  the  trees  bear  apples,  \ 


124  ?;o\iMON  ARTiHMETicr.  Sect.  YIIL 

bear  pea'clres,  rirrd  2fO  trees  bear  plums.     H&w  many  trees  are  there 
in  the  oi'chard?  Am.   lOG. 

1-7.  In  a  certain  school  |  of  the  pupils  study  arithmetic,  -^  study 
grammar,  j  geography,  ^^  lem-n-  to  write,  aad  15  learn  tc*  read. — 
How  many  pupils  in  the  school]  Ans.    120. 

18.  If  ai  certain  number  be  mcreas^d  by  J,  \,  j,,  and  ~  ofitself, 
and  6  more,  the  sum  will  be  doable  the  number.  Required — the 
numb€??  Ans.   1^, 

W.  A  manbemg  asL'ed  the  tFme,  answered,  "K  you  increase  it  bj 
^  of  itself,  rt  will  be  1 2  o'clock.     What  tim-e  Wc^s  it? 

Ans .  8  o'clock. 

20.  I  desire  to-know  the  time  by  knowing  that  ff  it  is  increased  by 
-^,  J,  and  i  of  itself,  it  will  be  half  past  12  o'clock. 

Ans.  G  o'clock. 

21.  "If  to  my  age  there  added  be,. 
One-half,  one-third,  and  three  times  three, 
The  whde  will  make  six  score  and  ten,. — ^ 

Pray  tell  111 Y  age  now  if  you  can."  Ans.  66  yea;rS', 


2,    DECIMAL    FRACTIONS^ 

Article  7.     Definitions,  &e. 

Rf.maick. — As  we  hav?  said  before,  if  a  uni£  is  divided  into  equal  parts  these 
parts  are  called  Fractio:\'s.     (Art.  1,  Obs.  4.) 

Obs.   1.     A  Decimai*  Fraction  is  one  in  wJdch   the  denominator 
is  1  with  any  number  of  ciphers  annexed;  as  ~-^^  t/o-'TFoV>  <^c. 

Note. TIae  word  decinyil  is  derived  frcwn  the  Latin  wcrd  decern,  which 

signifies  ten. 

Remark. — Decimal  Fractious  aregenoraliy  written  without  the  denominator 
being  expressed^  in  which  case  a  point  (  .  )  called  a  separatrix,  or  separ»/in^ 
point  is  placed  before  it  to  distinguish  it  froa^  whole  numbers.     Thus — 
.5,  .17,  .  479,  &c.  are  read  5  tjiiths,  17  huiidredtbs,   473    ihousaaiths,    &c. 
Hence — 

Obs.  2.   The  denominator  of  a  decimal  fraction  is  1 ,  idth  as  many 
ciphers  annexed  as  there  are'  figures  in  the  nmnerator,  or  deGimal.     Thus 

•  1=   tV 

•  ^  ''■'        10  0'  •  ^^  •  *-'  10000 

Whatis  the  denominator  of  .3?   .7?   ,19?   .54?   .1008?    .156? 
.98?   .2007?   .9? 
Suppose  it  v^'ere  required  to  write  -^l^  without  the  denominator. 

♦iSd  y cart. 


100  132 

•  ^^^  —  To  o"o• 
,4376=    '-''' 


f9Q4.7  12347 

.l^,U^I    lOO'O'O'O'* 

4.7fi8Q9  47  c  8  !)2 


Art  7.     > 


DGCMAL    FRACTtONf?, 


i25 


3 

a  0  0  0  0 


In  this  case  the  mmaerator  does  liot  contaia  as  many  figures  as  there 
are  ciphers  in  Hie  deoeminator,  but  this  is  remedied  by  writing  a  ci- 
pher before  the  4,  thus,  -04.  In  the  same  manner  jj^=  .006; 
=  .  0003,  &c.     Hence— 

To  write  decimals,  when  the  numerator  does aot  contain  as  many 
significant  figures  as  t^iei-^e  are  ciphers^in  the  denominator: 

Obs.  3.  Prcjix  ciphers  toihe  sh/njjicantjigures  of  the  mimtrcd,(yr,  un- 
til the  nmnher  of  iec'utwA pL^cc%  is  equal  to  the  nmnher  ofcipiners  mthe 
denommmior.     Thus — 


IS  written 


1.0  0  0 

Obs. 


-6 

.06 

-006 


1  II  »  4  M 

7o'o'o"a'o 
5^ 

I  0  0  0  (Ji)S 


IS  »'rittOil 


.     'Ooac 

.U0006 
.000006 
4.  From  this  it  appears  that  the  first  plac^^  at  the  right  of 
the  separatrix  is  called  tentms,  hece'U-ne  'aunit  h  divided  info  ten-equal 
p&Hs;  the  second  place  is  called  uv^i^n^BT^m  fi'om  ^itidhu/  -tenths  i'nt'<^ 
ien  equal  partM,  or  a  urdt  hito  u  hundred  eq^cd  paj*ts  ^  Ac.  This 
can  be  easily  shown  from  this  following 

DECLMAL  AUMERATIOSJ  TA^IJ^ 


73        - 


>i 


•^  g"  c  g  o  ,«::  5  —  c  5  — 


9.43 


S" 


read  6  teutlvs. 


is  read  Q  units.  43  liu-n- 
dredllis. 
o       1    "^  7 ^  s  read  12-uirits,  137  thou- 
sandth?, 
is  read  437  smfes,  1342  tea 

lii4>usandlhs. 
is  i^ea  *  1 2-45  units,   60044 

Im^ndiied  ll>3ue{iindths. 
is  r^-ad  <5C0C0  uiiits.,  4  ml- 
\      iftMiths. 

I  is  read  300   units,  '627.81 
f      ten  !milli©nt^:s. 
\  is  ]»ea(l  12  units,  0947321'$ 
4      h.undied  milliontlis. 
I  is    r^ad  200  umi^  -9  ibil- 
j     Jittntha, 


:    1 

4  3  7-0342::::: 
1  2  4  5 .  €  o  0  a  1  :  :  :  : 
0  0  0  0  -  0  o  o  o  0  4  :  :  : 

3  0  0  .  0  0  ^s  2  T  8  5  :  : 

12.  '8  ^S   47321    3: 

"* «  .0  .0  -  0  ^  ;0  .0  .0  0  .0  0  B 


l26  COMMON    ARITHMEnc.  Scct.  Vill. 

Obs.  5.  By  examining  this  table  attentively  ^ve  notice  the  fol- 
lowing coasiderati  ons :" 

1st.  Decimals  decrease  from  the  left  hand  towards  the  right  in  a  ten 
fold  ratio.  Thus,  .  4  is  only  4  tenths  ;  .  04  is  4  hundredths  ;  .004 
is  4  thousandths,  <t:c.  Therefore  conversely,  Decimals  increase  from 
the  right  hand  towards  the  lef-.  in  a.  tenfold  ratio.  Thus,  .  4  is  ten  times 
larger  than  .04  ;  .  04  is  ten  times  larger  than  .004  ;  &c.    Hence — 

Obs.  6.  Ennrg  removal  of  a  de:im,d  figure  to  the  right  decreases y 
and  every  r^noval  to  the  left  increases  its  value  ten  times. 

Remark. — The  pupil  will  perceive  that  decimals  ipcrease  and  decrease  in  the 
same  manner  as  Wiiaie  numlxTS. 

9,1.  The  value  of  every  figure,  whether  a  decimal,  or  cm  integer  is 
deter  mined  by  its  place  from  units.     There  fore — 

Obs.  7.  a.  Prefixing  one  cipher  to  a  decimal  decreases  its  value  ten 
times  ;  two  ciphers,  a  hundred,  times,  dx.  Thus,  .  3  =  j „  ;  .  03  = 
tI^;   .003  =  ^^V«,  &c.     But 

b.  Annexinr/  a  cijpher  to  a  decimal,  however,  does  not  alter  its  value. 
Thus,  .7  ="-1..;  .70-  ^L  =  _T_;  _ZJLiL  =  ,v^,  as  betbre.— 
Hence — 

Obs.  8.  Deimals  of  difi'ereni  denominators  may  he  reduced  to  a 
common  denominator,  hy  annexing  ciphers  witil  the  nimiher  of  decimal 
places  in  each  ((re  equal.  Tluis,  .4,  .00,  .037,  are  equal  to  .400, 
.060,  .037,  &c.     Aho— 

Obs.  9.  Whole  n:irn')er3  may  he  reduced  to  decimals  hy  annexing 
ciphers.  Thus,  16==  V/»  V^V^  ^^  l^J  tenths,  1600  hundredth?. 
(Art.  3,  Obs.  7.) 

Remark. — When  the  w!  o-lc  number  U)i5t5  reduced  is  written  without  the  de- 
nominator, it  is  best  to  place  ihe  seiirirjirix  before  the  ciphers.  Tiius,  1G=  16 
.0=  iG.OO.&c. 


"VVJiat  rjre  Fractions?  VV'iiat  isaDicimjl  Fra-^tiou?  Fro  ri  what  is  the  term 
decimal  derived  ?  How  artf  decun^il  fractioiis  generally  written?  What  do  wo 
iise  ill  this  case?  Wluri- is  tliis  point  p'ac-ed?  Why?  VViiat  is  the  denomi- 
nator of  a  deiMmal  frav^'.ion?  Ijow  do  we  write  the  decinial,  when  the  iininer- 
ator  does  not  contain  as  many  ^igni^ca»l  Hf/urcs  a.*  theie  are  <  i,)ht'rs  in  the  de- 
nominator? What  is  tl>e  lirst.  place  i*l  ti)e  right  of  ihe  si-panlrix  called?  Wliy? 
The  second  place?  Why?  How  do  dicin)a!s  decrease?  How  increase? 
What  inference  is  dednccd  from  Hiis?  \\  h.at  is  the  difference  between  the  in- 
crease and  decrease  of  decimals  and  vvIjo  e  jn^mbers?  How  is  the  value  of 
every  f^gfuredctermiii;^  I :  Wh  -t  elFect  does  it  liave  to  j>re(lx  a  cipher  to  a  deci- 
mal? Two  ciphers?  What  efTrCt  does  it  have  to  annex  a  cipher  to  a  decimal? 
How  do  we  rt'dnced'  cimaKs  of  dilfe  eat  denominators  to  a  comm.oa  denomina- 
tor? Hov^  »nay  whole  nnmbers  he  reduced  to  decimals?  When  the  whole 
number  is  written  vvilhout  ihe  denominator,  how  do  we  proceea?  How  may 
whole  numbers  and  aecirrjaU  be  wriUetv  tojrelher?  Whatare  such  expressious 
called? 


Art.    7.  lEC'LMAf.    FRACTIONS.  127 

^  3(3.  Whole  numbers  and  decimals  may  he  wntim  together  hy  ;  la- 
cing the  seperating  point  between  them.  'J  lius,  9  units,  9  millionths  is 
written  9  .000009. 

a.  A  whole  number  and  decimal   loritten  together  is  called  a  Mixed 
Number.     Thus,   4  .7,  6  .08,  and  4.027  are  mixed  numbers. 

4th    The  units  place  is  at  the  right  of  whole  numbers,  and  at  the  left 
of  decimals.     Hence — 

The  effect  of  annexing  or  prefixing  ciphers  to  decimals,  is  tJie  reverse 
of  annexing  or  prefixing  them  to  whole  numbers. 

5th.  The  name  of  the  order  of  the  rig  Jit  hiand  figure  of  the  decimal 
is  given  to  tlie  whole.     Hence — 

To  read  decimals: 

Obs.  10.  Read  as  ill  whole  numbers,  and  to  the  right  hand  figure 
add  the  name  of  its  order. 

Note. — When  mixed  numhers  are  read,  it  is  preferable  to  pl;u'e  the  word 
«ni/s  aft*  r  the  whole  number,  to  prevent  ambijruiiy.  Thu-,  400.  0)16  woul4 
be  read  by  many  pupils  416  ten  tholl^:andli1s.  But  by  reading  it  400  units,  and 
16  ten  thousandths,  all  ambiguity  is  removed. 

Remark  1. — Expre.ssiug  decimals  by  words  is  culled  Numeration'  of  Deci- 
mals. 

Read  the  following  decimals: 

1.  .5.       \b.     12.00021.     i  9.   1200.000000016.^13.       4.07. 

2.  .27.      io.   217.1G2345.    jlO.        78.004.  114.        8-167. 

3.  1.081.   J7.  400.0000096  111.      120.3.  ;15.      11.4092. 

4.  9. 3906. :8.  210. 3000009.112.     100. 0C6.  ;  16.  700  .0075. 

Rkmaric  2. — Expressing  decimals  by  figures  is  called  Notation  o?  Deci- 
mals. 

To  write  decimals : 

Obs.  11.  Write  each  figure  in  the  order  in  vhich  it  belongs^  and 
place  a  ciplier  in  all  vamnf  orders. 

Write  the  fniclionnl  parts  of  the  fol1ov.-in:;-  tmrnliers  in  de^^i- 
mals  : 


Where  is  the  niiits  pluce  in  whole  numbers'  In  decimals?  What  is  the  dif- 
ference in  the  eflect  of  annexing  and  prefixing  cipliers  to  wisoie  numbers,  and 
4g  decimals?  What  name  is  given  to  the  d^'cirnul?  liow  th.  u  do  we  read  deci- 
mal.s?  How  is  it  best  fo  read  mixed  numbers?  Wb.yso?  What  is- numeration 
of  4<-'^-i">a^s?     What  is  uo-tation  of  decimals?     How  do  we  wriio  decimals? 


128 

1.* 

2. 
3. 
4. 
5. 
6. 
7. 
8. 
9. 
10. 


COMMON    ARITHMETIC, 


Sect.  VIII 


'  10' 

27--- 
130y|fo, 

'^" 1  0  0  0 
*^l  0  0  0  0' 
■1 '-'  1  n  n  n  n 


Ans.  7.3. 


Ans.  14.09. 


' 'To  0  0  0  00  0* 

1  UUOy  Q-Q-g- Q- Q-Q-g-jj- 0-. 


49- 


11. 
12. 
13. 
14. 

15. 
16. 
17. 

18. 
19. 

20. 


l^^To'o'o'' 

169- 
463 


l_8_ 

fo  0* 
3  3  3 


1  0' 

1 
1  0  0' 


3  C 
"0"0  0" 


1 

400 
78A. 
200^0^077- 
60xHo.  ^ 

147Dyo-^O^O-0^. 


Write  the  following  expr  ssions  in  decimals 


21.  7  teatlis 

22.  4  hundretlis. 

23.  14  thousandths. 

24.  6  millionths. 

25.  1 8  hundredths. 

26.  36  thousandths. 

27.  9  tenths. 

28.  112  ten  thousandths. 

29.  4002  ten  millionths. 

30.  3  units.  4  thousandths. 


31.  7  units,  12bilIionths. 

32.  1 4  units,  2  thousandths. 

33.  612  units,  25  hundredths. 

34.  406  units,  406  thousanths. 

35.  400  units,  6  thousandths. 

36.  1000  units,  1  millionth. 

37.  79  units, 2001  ten  thousandths 

38.  976  ten  biilionths. 

39.  1001  hundred  trillionths. 

40.  100  units,  111  millionths. 


Note. — The  reading  and  writing  of  decimals  is  of  great  importance,  and  the 
pupil  should  be  exercised  at  it  until  it  is  perfectly  familiar  to  him. 

Art.  8.     Federal  Monev, 

Obs.  1.  Federal  Money  is  the  national  currency  of  the  United 
K5TATES,  as  established  by  Congress,  Angust  8Lh,  1776.  The  de- 
nominations are  Mill,  Cent,  Dime,  Dollar  and  Eagle. 

table. 

10  mills  (??i.) make  1   cent marked  c^ 

10  cents **  1   dime "  d. 

lOdimes •'  1   dollar ''doll. or^ 

10  dollars **  1  eagle ''        K 

Note. — This  character  ($)  may  be  regarded  as  a  contraction  of  U.  S.,  and 
signifies  United  States  money. 

1st.  The  Uaffle  is  a  gold  coin,  and  contains  10  pennyweights,   18 


What  is  Federal  Money?     What  are  the  denominations?     Repeat  the  table? 
What  are  the  national  coin  of  the  United  States? 


AVt.  6.  ^  -fEDERAL   MONEY.  V2% 

trains  =—  258  grains  of  standard   gold.*     Besides  'the   eagle,    we 
ave  the   half-eagle  and  quarter-rngk.,   which  are  gold  coins,    and 
'their  value  and  weight  accordingly, 

2d.  The  Dollar  is  a  silver  coin,  and  contains  1 7  penny  weights,  A~ 
;grains=  41 2|^  grains  of  standard  silver. f  Besides  the  dollar  there 
are  half-dollars  y  quarter -dollars,  dimes,  and  half -dimes,  Avhich  are  sil- 
ver coins,  and  their  value  and  weight  accordingly. 

3d.  The  Cent  is  a  copper  coin,  and  contains  7  pennyweights  =  168 
grains  of  pure  copper. J;     Tlie  half-cent  accol-dingly. 

Jfiils  are  only  imaginary,  and  are  not  coined. 

^Obs.  3.     Pure  gold  is  supposed  to  be  divided  into  24  equal  parts 

called  carats,  and  its  fineness  depends  on  the  number  of  parts  of  some 

baser   metal  called  alloy  that  it  contains.     Thus,  if  it  contains  20 

parts  pure  gold,  and  4  parts  alloy,  it  is  said  te  be  20  carats   fine^;  if 

'it  contains  6  parts  of  alldy,  it  is  said  to  be  18  carats  fine. 

Obs.  4.  Previotis  to  1837,  the  standard  for  gold  was  22  parts  of 
,pure  metal,  to  2  parts  of  alloy,  and  it  was  said  to  be  '22  Carats  fine. 
Now  it  is  21,|  earats  fine. 

Obs.  5.     By  Act  of  Congress,  ■■  1 837,  the  legal  standard  for  gold 

and  silver  coins  in  the-UNiTED  States,  is  900  j^ar^s  of  joure  mftal, 

hy  iceight,  to  100  parts  alloy.     The  alloy  of  gold  coin  is  composed  of 

^s  ilv  r  and  copper,  the  silver  not  to  exceed  copper  in  weight,    '?The 

alloy  of  silver  corns  is  pure  copper. 

Obs.  6.  Acoown^  in  /Ae  United  States  ar3  usually  kept  in  dol- 
lars,  cents,  and  mills  ;  eagles  being  expresssed  as  dollars,  and  dimes 
as  cents.  Thus,  instead  of  saying  2  eagles,  6  dollars,  we  say  25 
dollars;  and  instead  of  saying  7  dimes,  5  cents,  we  say  75 -cgnts, 
&c.  Five  mills  are  often  called  ^|'^.  cent ;  thus  12  cents,  Senilis, 
are  generally  called  1 2-J  cents. 

Remark. — It  will  be  perceived  from  the  table  that  the  deneminStionsHaf  Fed- 
eral iMoney  increase  and  decrease  in  a  ten  fold  rai!w,iti  the  same  manner  as 

,,.  What  is  the  eagle?  Its  weight?  What  other  gold  coins  have  we?  W4>at  is 
Hhe  dollar?  Its  weight?  What  other  silver  coins  have  we?  What  is  the  cent? 
'Its  weight?  What  are  mills?  How  is  pure  gold  divided?  Upon  what  does  its 
fineness  depend?  If  it  contains  4  parts  alloy,  how  fine  is  it?  Whal-wtis  the 
•standard  for  gold  previous  to  1837?  What  is  it  now?  What  is  the  legal  stan- 
■'dard  for  gold  and  silver  corns  in  the  United  Slates,  by  Act  of  Congress- of '1837? 
Of  what  is  the  alloy  of  gold  coins  composed?  Of  silver  coins?  How  are  'f-.c- 
connls  in  the  United  States  gentTiliy  kept?  How  are  eagles  expressed? — 
Dimes?  What  do  we  say  insteni  of  2  eagles,  5  dollars?  7  din>es,  &i?(J  5  cents? 
What  are  5  mills  often  called?  IJow  isl2  e»nts,  5  mills, generally  read?  Ujow 
do  the  denominations  of  Feder;'!  Money  increase  and  decrease? 

*  Ka,'!o^coinel  before  July  31«t,  ;83-I,  -oatain  11  pennvweislits,  B  grains  =  £70  grains  of 
stancla\>l  gold.     These  are  worth  SIO  .  <505.     Half  and  quarter  eaghsa'-cordin'.riv- 
t  'nie.*Iollar  originally  contained  17  pennyweights,  8  grains  =  419  grains  of  si.julard  sUvcr. 
X  Tliecent  originally  contained  264  crrains  of  pure  coppc". 

7a 


130  COMMON  AtLirHMEtnc,  Sect.  Till. 

whde  numbers.     Hence — the  dollar  being  regarded  as  the  wait,  and  the  ce»t9 
and  Bvill®as  the  fractional  parts  o-f  the  uuH  or  dotkir,  it  foMows  tJiat 

Obs.  7.  u4ll  operations  in  Federal  Monep  can  he  'performed  pre- 
cisely as  in  decimal  fractions,  the  dollar  being  regarded  as  the  unitf 
cet^s  as  tenths  and  hundredtltSy  (because  100  cents  make  a  dollar)  arid 
the  mills  as  thousandths;  the  separating  point  being  placed  between  the 
dollars  and  cents. 

Obs.  8.  As  cents  are  so  niatij  liundredtlis  of  a  dollar,  (it  taking 
100  to-  make  a  dollar,)  it  follows  that  they  must  occupy  the  first  two 
places  at  the  right  of  dollars,  and  if  ihe  cents  are  less  than  10,  a  ci- 
pher must  be  placed  at  the  left,  or  in  ihe  tenth  place.     Thus,  1 8  dollars, 

5  cents  is  writtea  ^18  .05. 

likewise,  as  mills  are  so  Diaiiv  thousandths  of  a  ddilar,  (it  takii^g 
1000  to  m^ke  b  dollar,)  they  must  occupy  the  third  place  at  the  right 
of  dollars-,  and  if  no  cents  are  given,  the  place  of  cents  must  be  sup- 
j^iedwit/i  ciphers-. 

Thus,  9  cts^  is  written  .Ot;  J8  dolls.  6  ets.  7  m.,  is  wri  ten 
^18  .067  ;  15 dolls.  37  cla.5  m.,  is  written  $t5.m&,  or   ^16^37| ; 

6  ddls..  9   m.,    is   written    $6  .009  ^  and  7  m.,  is  written  ^0.007  ; 
&c. 

Remark. — Busi^aess  m-en  &ften:  write  cents  aa  tlie  fractional  paKtoof  Efdo'llaf.- 
Tkus,  they  write  $10  .  46,  $10_y_.  &lc.  [ 

To' read  any  sum  in  Federal  Money  :- 

Ot)s.  9'.  Call  the  f glares  at  the  left  of  the  separating  point  dolljccrs^r 
the  first  two  places  at  the  right  cents,  and  the  third  place'  ai  the  righ 
mills  ;  the  other  2^laces  at  the'  righd'  are  decimals  of  a  mill. 

Thi?«,  %m  .275  J6  is  read  5Q  dolls,  27  cts,.  5  m.,.asd  JS:  hundredths 
etf"  ami!].. 

I^OTE, — Th-e  decimals  at  the  right  of  mHls-arc  seldom  counted,  th«  mills- 
being  sufficiently  exact  for  all  business  calculations.  Some  however,  reckon 
iu' this  way — if  the  ni ills-exceed  5,  they  count  another  cert;  if  they  are  less 
than  5  they  reject  them  entirely  ;  5-m.  they  count  k  a  eentv 


How  c-an  operations  in  Federal  Money  be  perftjrmedT  What  is  the  dollar 
regarded?  The  cents?  Why?  The  mills?  Where  is  the  separating  point 
pi'aced?'  By  what  are  the  first  two  placesatthe  righ^t  of  the  separating  point  oc- 
carpied?'If  the  cents  are  less  than  10  how  do  we  proceed?  Wiiat  place  do  mill* 
©ccupy?  If  no  cents  are  given  wl>at  must  be  done?  How  do  business  men  of- 
ten write  cents?  How  do  we  read  any  sum  in  Federal  Money?  Why  are  not 
the  decimals' a-t  the  riglU  of  mills  generalty  reckoned?  How  do  sonve  reckon. 
sullisZ' 


Art.  9. 


FEDERAL    MONEf. 


131 


Read  the  following  suras  in  Federal  Money  : 


1. 
2. 
3. 
4. 
5. 
6. 
7. 
8. 
9. 
10. 


ao.06. 
^1  . 703. 
$17  .6954. 
$243,009. 
$78  .  805. 
$20;!.  73021. 

$14.  12}. 
$100.50. 
M086.375. 
$12.1875. 


11.  $70,006. 

12.  $32671.09999. 


13. 

$14,007. 

14. 

$0,001. 

15. 

$0,158. 

16. 

$12.70. 

17. 

$9.37f 

18. 

$14.18|. 

19. 

$0.56^. 

20. 

$100,001. 

Obs.   10.     We  write  sums  in  Federal  Money  according  to  (Art.  7  > 
Obs.  11.)  dollars  occupying  the  place  of  whole  numbers,  cents   the 
place  of  teMks  and  hundredths,  and  mills  the  place  of  thousandths. 

Thus,  $  1 8 .  37  cts.,  5  m.,  is  written  $18. 375,  &c. 

Write  the  following  sums  in  Federal  Money  : 


1.  27  dolls.  49  cts.  7  m. 

2.  18  dolls.  18  cts.  8  m. 

3.  20  dolls.  27  cts.  6  m. 

4.  97  cts.  4  m.;  68  cts. 
6.  78  cts.  9  m.;  47  cts.  1  m, 

6.  1007  dolls.  87  cts.  6  m. 

7.  947  dolls.  69  cts.  3  m. 

8.  7  dolls.  5  cts.  9  m. 

9.  18  dolls.  6  cts.  3  m. 

10.  19  dolls.  9  cts.   1  m. 

11.  34  dolls.  7  cts.  8  m. 

12.  19  dolls.  2  cts.  3  m. 

13.  10  dolls.  9  m.;  4  cts. 

14.  74  dolls.  3  cts.  7  m, 

15.  143  dolls.  9  m. 

16.  9  m.  and   12  hundredths  of 

a  mill. 

Article,  9.     Rkduotion  of  Federal  Money  and  Decimal  Frac- 
tions. 

Obs.  1.  As  10  mills  make  a  cent,  and  100  cents  make  a  dollar, 
it  follows  that 

a.  Dollars  are  reduced  to  cents  by  annexing  two  ciphers,  and  to  mills 
by  annexing  three  ciphers.  In  either  case  we  remove  the  sign  of  dol- 
lars ($). 


17.  1  doll.  Im.;  3  m. 

18.  3  cts.  2  m.;  7  m. 

19.  129  dolls.  8  cts.   1  m. 

20.  6  m.;7  m.;  4  m. 

21.  8  m.;  1  ct.  1  m. 

i:2.  200  dolls.   14  cts.  2  m. 

23.  93|cts.;  4  m. 

24.  75  cts.  1  m. 

25.  1  doll.  I  ct.  1  m. 

26.  20  dolls.  2  m, 

27.  79  dolls.  4  m. 

28.  1  m.;  9  m.  5  tenths  of  a  mill, 

29.  25  hundredths  of  a  mill. 

30.  100   dolls.   1    ct.    75   hun- 
dredths of  a  mill. 


How  do  we  write  sums  \i\  Federal  Money?    How  are  dollars  reduced  to 
ceaU? 


132  COMMON  a!rithM£tic.  Scct.  Vlll. 

Dollars  and  cents  are  reduced  to  cents,  and  dollars,  cents,  and  mills 
are  reduced  to  mills  by  erasing  ike  separating  jcoint,  and  the  sign  qf 
dollars. 

h.  Cents  are  reduced  to  dollars  by  jpohiting  c^  two  figures  at  the  rigU 
and  prefixing  the  sign  of  dollars.  {Jents  are  reduced  to  mills  by  an- 
Tiexing  a  cipher. 

c.  Jills  are  reduced  to  cents  by , pointing  of  three  figures  at  the  right 
and  prefixing  the  sign  of  dollars. 

EXERCISES    FOR    THE    SLATE-. 

1.  Reduce  $17  to  cents.  Anr.   1700  cts. 

2.  Reduce  $34  to  mills. 

3^.  Reduce  $  1 8 .  73  to  cents. 

4.  Reduce  $25,645  to  mills. 

5.  Reduce"  $1.01  to  mills. 

6.  Reduce  $4  .20  to  cents. 

7.  Reduce  $27  .06  to  mills. 

8.  Reduce  $0  .479  to  mills.  '    ^ 

9.  Reduce  $12.06  to  cents.  '' 

10.  Reduce  478  cents  to  dollars. 

11.  Reduce  164  cents  to  mills. 

12.  Reduce  2080  cents  to  dollars-, 

13.  Reduce  14000  cents  to  mills. 

14.  Reduce  120  mills  to  c:nts. 

15.  Reduce  14000  mills  to  dollars. 

16.  Reduce  1785  mills  to  cents. 

17.  Reduce  800  cents  to  dollars. 

18.  Reduce  1768  mills  to  dollars. 

19.  Reduce  12435  mills  to  cents  and  dollars-. 

Reduction  of  Declmal  Fractions. 

Obs.  2.  The  learner  will  remember  that  inull  cases  of  deduction, 
whether  of  Common  or  Decimal  Fractions,  the  value  of  the  given  number 
must  not  be  altered.     (Art.  3,  Obs.  1.  and  Art.  6,  Case  5,  Rem.) 

Case  1 .      To  change  a  Decimal  to  a  Common  Fraction. 

Remark. — As  we  have  said  before,  the  denominator  of  a  decimal  fraction  is 
1  with  as  many  ciphers  aiuiexed  as  there  are  places  of  decimals  in  the  numera- 
tor.    (Art.  7,  Obs.  2.) 

Hence — To  change  a  decimal  to  a  common  fraction  :  ^ 

To  mills?  What  is  necessary  iu  these  cases?  How  are  dollars  and  cents  re- 
duced to  cents?  How  are  dollars,  cents,  and  mills  reduced  to  mills?  How 
are  cents  reduced  to  dollars?  To  mills?  How  are  mills  reduced  to  cents.? 
To  dollars?  What  must  be  observed  in  all  cases  of  reduction?  What  is  the 
denominator  of  a  decimal  fraction?  How  then  do  we  change  a  decimal  to  a 
common  fraction?  ...,*.x   .- 


Art.   9.  REDUCTION    Of    DECIMAL   fRA€TIONS.  l''3'3 

Ohs.  3.  Erase  the  decimal  point,  and  supply  the  denominator  of 
the  decimal : 

Note.' — It  is  gefnerallycestomary,  after  the  decimal  is  reduced  to  a  ebminon 
fraction,  to  reduce  it  to  its  lowest  terms.  (Art.  3)  Obs.  3.)  This  is  the  case 
with  the  following  examples. 

Ex,  1,  Change  .25  to  a  conwiion  fraction*  Ans.  y~  =  \. 

2.  Change  .  125  to  a  common  fraction,  Ans.  \, 

3.  Change  . 06^25  to  a  common  fraction*  Ans.  ~g, 

4.  Change  .  7425  to  a  common  fraction.  Ans.  |§  J, 
•5,  Change  .75  to  a  common  friction,  Ans.  J« 
6.  Change  .  875  to  a  common  fraction,  ■  Ans.  |-, 
9'.  Change  ^  5625  to  a  common  fraction.  Ans.  y\k 

8,  Change  .  9375  to  a  common  fraction,  Ans.  —i 

9.  Change  .333  to  a  common  fraction.  Ans.  yoVo-. 

10.  Change  .99  to  a  common  fraction.  Ans.  y—g-^ 

11.  Change  .1246  to  a  common  fraction.  Ans.  sVo^-g-. 

12.  Change  .03125  to  a  common  fraction.  Ans.  —, 
"Case  2.     To  change  a  Common  Fraction  to  a  Decimal. 

Note. — This  case  is  exactly  the  revefseof  Ihepreceding  one,  and  each  proves 
the  other. 

Ex.   1,   Change  |  to  a  decirftal, 

Operalloh. 
Annexing  a  cipher  to  our  numerator,  we  make  it  20  6)2.0 

tenths:    20   tenths   divided  by    5  (the  denominator)  

Equals  4  tentlis.  Alls.   .  4 

The  correctness  of  this  operation  may  be  shown  as  follows: 

Annexing  a  cipher  to  the  numerator,  is,  in  reality  multiplying  it 
by  10;  (Sect.  IV,  Art.  4,  Case  1,  Rem,)  btit  multiplying  the  numer- 
ator is  multiplying  the  value  of  the  fraction;  (Art,  2,  Obs,  1.) 
hence,  the  quotient  (after  dividing  the  numerator  by  the  denomina- 
tor,) is  10  tim^s  too  large,  and  we  must  therefore  divide  it  by  10  to 
obtain  a  correct  result.  This  is  done  by  simply  pointing  oflf  a  figure 
at  the  right,  or  placing  the  decimal  point  betbre  it.  (SeTit.  V,  Art, 
4,  Obs.  1.) 

By  the  same  course  of  reasoning,  it  follows  that  if  we  annex  two 

—————— — . . , —^ ■ — ■• 

What  relation  does  Case  1,  and  Case  2  bear  to  each  other?  Explain  why  the 
Operation  of  Ex.  1  is  correct.  If  we  annex  two  ciphers  to  the  numerator, 
what  ofTect  does  it  have  on  the  quotient?  If  We  annex  thlfee  ciphers,  what  is 
the  effect?    Why  bo? 


134  CoMAfoJf    ARITHMETIC*  Sect.  VIII. 

ciphers  to  the  numerator  we  multiply  it  by  100,  and  must  point  ofif 
two  decimals  in  the  result ;  if  we  annex  three  ciphers  to  the  numer- 
ator, we  multiply  it  by  1000,  and  must  point  oif  three  decimals  in 
the  result,  &c,     Hence-^^^ 

To  change  a  common  fraction  to  a  decimal: 

Obs.  4,  Annex  ciphers  to  ike  numerator,  and  divide  hy  the  denom' 
inator, pointing  off  a  decimal  in  the  quotient  for  every  cipher  annexed; 
and  if  there  is  not  a  S2(fficient  numher  of  figures  in  the  quotient,  supply 
the  deficiency  hy  prefixing  ciphers. 

a.  From  this  and  the  last  case,  the  learner  can  often  form  rules  for 
contracting  the  operation-  in  multiplication  and  division.  Thus»  to 
multiply  by  *  25,  he  can  divide  by  4,  because  .  25  ==  ^  ;  to  multiply 
by  .  125,  diade  by  8,  because  Alb  =z\  ;  (fee;  and  conversely  to 
divide  by  .  25,  multiply  by  4,  and  to  divide  by  .  125,  multiply  by  8, 
&c.,  for  the  same  reason, 

Again,  to  multiply  by  .  875,  we  multiply  by  7,  and  divide  by  8, be 
cause  .875=  J,  &c,;  and  conversely,  to  divide  by  .875  we  multiply 
by  8,  and  divide  by  7  for  the  same  reason, 

But  when  our  multiplier  is  a  whole  number  (and  we  contract  the 
operation,)  we  must  annex  to  the  product  1,2,  3,  &c,  ciphers,  be- 
cause our  multiplier  is  a  part  of  10,  100,  1000,  &c,;  and  conversely 
when  our  divisor  is  integral,  we  must  point  off  from  the  right  1,  2,  3, 
(fee,  figures  for  the  same  reason.  The  figures  cut  off  however, 
must  be  divided  by  our  multiplier,  because  these  are  the  remainder, 
and  are  as  many  times  too  large  as  there  are  units  in  this  multiplier, 
(Sect.  VI,  Art  2,  Rem,  under  Ex,  22  and  23,) 

We  can  also  abridge  the  operation  when  the  multiplier  or  divisor 
is  partly  integral,  and  partly  decimal ;  thus  to  multiply  or  divide  by 
1 .25,  we  proceed  exactly  as  if  it  were  125.  In  such  cases  we  only 
annex  as  many  ciphers,  or  point  off  as  many  decimals  as  there  are 
integral  numbers  in  the  multiplier  or  divisor.  When  there  are  deci- 
mals in  the  dividend  or  multiplicand,  we  may  proceed  with  them  ac- 
cording to  the  rules  in  multiplication  and  division  of  decimals. 

It  may  be  proper  to  remark  that  the  subject  of  abbreviations  can 
only  be  useful  to  a  person  well  acquainted  with  the    properties   and 

How  then  do  we  change  a  common  fraction  to  decimal?  To  what  pdrticu- 
lar  object  can  these  two  cases  be  applied?  €Hve  an  example  or  two  illustrating 
this  point?  When  our  multiplier  is  a  whole  number,  how  do  we  proceed? 
Why?  When  the  divisor  is  integral,  how  do  we  proceed?  Why?  v\  hat 
must  be  done  with  the  figures  cut  off?  Why?  When  the  fnuUiplier  or  divisor 
ia  a  mixed  number,  how  do  we  proceed?  VVh^t  tiuist  be  observed  in  such 
cases?  When  there  are  decimals  in  thts  divi;if  nJ  or  maltipiicHiid,  how  do  we 
proceed?  What  is  necessary  iu  order  to  tnuke  Uic  subject  of  abbreviations 
UMful  to  a.  person? 


Art.    10.    AD   ITioN    AKf)  SUEtRAcTiCN  OF   DeCLMaLS,   LC.  135 

tdsitions  of  numbers,*  r(,nd  with  such  a  few  limts  are  generally  suf'^ 
ticient.  We  will  therefore  only  present  the  two  following  proposi- 
tions, which  are  so  evident  that  we  omit  the  demonstration  : 

1,  ^'  tve  ivish  to  divide  one  niimhev  by  another^  it  will  produce  the 
'  same  effect  to  multiply  the  dividend  by  the  reciprocal  of  the  divisor. — 

And 

2.  If  vie  wish  to  multiply  one  numher  by  another.,  we  can  obtain  the 
same  result  by  dividing  the  multiplicand  by  the  reciprocal  of  the  multi- 
plier. 


2, 

Change  |  to  decimal, 

Ans,    .75 

3. 

Change  j  to  a  decimal. 

Ans.   .6, 

4. 

Change  -}o  to  a  decimal, 

Ans.   .05, 

b. 

Change  -fls  to  a  decimal. 

Ans.   .048, 

6, 

Change  |  to  a  decimfth 

Ans,    .375, 

7, 

Change  I  to  a  decimal. 

Ans,   1  ,  25, 

8, 

Change  --j  to  a  decimal. 

Ans,   ,36. 

9, 

Change  g|j  to  a  decimal, 

Ans,  ,008. 

10. 

Change  toVo  ^o a  decimal. 

Ans,  ,007. 

11, 

Change  |  to  a  decimal. 

Ans,   ,333333+, 

12. 

Change  ^y -j  to  a  decimal. 

Ans.   .468468+, 

In  the  last  two  examples,  it  will  be  perceived  that  the  decimal  will 
be  continually  repeated,  the  same  remainder  occuring  after  a  certain 
number  of  divisions, 

Obs,  5.  Decimals  which  consist  of  the  same  figure  or  set  of  figures 
continually  repeatedy  are  called  Repeating  Decimals,  Circulating 
Decimals,  or  Repetends, 

Article  10.     Addition  and  Subtraction  of  Decimals  and  Fed- 
eral Money. 

Obs.  1.  Decimal  Fractions  can  be  added,  subtracted,  multiplied, 
or  divided,  the  same  as  whole  nwnhers.  The  only  difficulty  is  in 
knowing  where  to  place  the  separating  point. 

Obs.  2.  As  Federal  Money  is  based  upon  the  decimal  nota- 
tion, ;t  is  evident  that  it  is  subject  to  the  same  laws  as  decimal  frac- 
tions, and  therefore  the  same  rules  are  applicable  to  both. 

Ex.  1,  A  man  has  several  accounts  to  collect;  the  first  is  $12;  the 

If  we  wish  to  divide  one  number  by  another,  by  what  other  way  can  we  ob- 
tain the  same  rosutt?  If  we  wish  to  multiply  one  number  by  another,  by 
What  other  way  can  we  obtain  the  sam«  result?  Demonstrate  these  proposi- 
tions. What  are  circulating  decimals  or  repetends  What  is  the  only  difficul- 
ty .in,ithe  addition,  &.C.,  of  decimals?  Upon  what  is  Federal  Money  based? 
What  is  the  inference  deduced  from  thi»? 

*  The  precedius  rules  and  prin^ipks  cxijiain  all  that  Uneccssvy  to  understand  this  point. 


14f6  •^comm6n  AkiTHmETrc.  Sect.  VIH. 

second,  i9,375;  the  third,   86,08;  and  the  fourih,   $5,403.     Re- 
quired— the  araount  of  the  whole* 

Oi^eration. 

In  this  example  -^ve  place  dollars  tindei*  dol-  Si 2. 000 
lars,   cents  u;. dor  cents,  and  mills  under  mills,  9.375 

ftnd  add  in  every  respect  as  in  Addition  of  6-.  080 

Simple  Numbers.     (Sect.  II.  General  Rule.)  #.  403-       v 

Ins.  $32,858 

2.  A  man  has  lii  one  bag  2.3  bushels  of  oats,  in  another  3. 161 
'bushels,  in  another  1.34  bushels,  and  in  a  box  14. 132  bushels. 
llt)w  much  has  Ire  4n  all? 

Operation. 

in  this  example,  we  place  whole  iimii-  2.300 

*)efs  under  whole  numbers,  tenths  under  3.161 

tenths,    hundredths    under    Imndredtiis,  1.340 

tfec,  and  add  as  ustial.  44. 132 


Ans.  2a.  933  bushels. 

Note — In  both  these  examples,  it  will  be  perceived  that  we  reduced  all  the 
decimals  to  the  same  denbmi nations  by  annexing  ciphers.  This  however,  is  not 
Absolutely  necessary  if  the  pupil  places  all  the  figures  of  the  same  orderdirect^ 
Ly  under  each  other. 

That  these  operations  are  correct,  may  be  shown  as  follows  •: 

3.  Add  ^together  5.8^6  and  7.9. 

Operation. 
5^.846     ^,,      ^5.846  =  5^VVo 

7Q  '       i7Q        7—^        ^     no  0.900    _i      8  4  6 

'••^  (/.a        — ./jj,.      jj,   —  looo'iooo     I*  To"  o  o" 

=  IJIJ  =  I^V/f-     7  +  5  =±  12,  and 

Ans.  13.746  ly^V^  =  l3,VoV      But  ^\Vo   =   -746. 

Hence^l3/oro  =  13.746,  as  before. 

"the  same  reasoning  will  apply  to  Federal  Money,  or  to  any  niim^ 
bgr  of  decimals. 

4.  A  man  having  $18.25,  spent  $9,625.  How  much  had  he 
left? 

Operation. 

In  this  example,  \^q   write  dollars  under  $18,250 

dollars,  cents   under  cents,  and  mills  under  9.625 

mills,  and  subtract  precisely  as  in  Subtraction  

•of  Simple  Numbers.  (Sect.  III.  General  Rule.)     Ans.   8.  625 

Explain  why  the  operations  given  are  correct. 


Art.    10.    REDUCTION    AND    SUBTRACTION    OF    f  ECIMALS,    &C.       137 

jg-QTE. — As  there  are  no  mills  in  the  minuend,  we  write  a  cipher  inthe  mi  Is' 
place  ;  but  this  is  not  necsssary,  as  we  always  proceed  as  if  the  cipher  was 
there,  whether  we  express  it  or  not. 

5.  A  man  having  18.4  yards  of  cloth,  sold  12.694  yards.  How 
much  had  he  left? 

Oj^eratlon. 

In  this  example,   we  write   whole  numbers  1 8 .  4000 

under  whole  numbers,  tenths  under  tenths,  12.694 

&c.,  and  subtract  as  usual.  

Ans.  5.706  y'ds. 

It  will  be  perceived  that  we  annex  ciphers  to  the  mmuend  until 
the  number  of  decimal  places  are  equal  to  those  in  the  subtrahend. 

The  same  example  may  also  be  solved  thus  : 
■to  A        1Q4       io4     iR-'f'"    1 7lli^   ■  1 7llii 12A?JL  = 

1  O  .  4^  =    1  OjQ.       1  O  Jo"     I  O-YQQ-yi   ^'lOOO*         *'iooo  *'*'1000 

12.694  =  12y|J±.  S^VoV  =  5.  706. 

This  shows  the  operation  to  be  correct.  The  same  method  of 
reasoning  w^ill  apply  to  either  Decimals  or  Federal  Money. 

Obs.  3.  From  the  preceding  operations,  we  notice  two  conside- 
rations : , 

1st.  The  separating  points,  both  in  the  given  numlers  and  in  the 
resulty  all  fall  directly  under  each  other, 

2nd.  In  Subtraction,  that  number  is  made  the  minuend,  in  which 
the  whole  number  is  the  largest,  without  reference  to  the  Decimals, 
because  the  Decimal  is  only  a  part  of  unity,  und  therefore  merely 
a  fraction.     Hence — 

To  add  or  subtract  Decimals  or  Federal  Money,  we  have  this 

Rule. — I.  Write  the  numbers  so  that  the  same  orders,  and  also 
the  separating  points,  shall  all  fall  directly  under  each  other.  (Sect. 
II.  Art.  2.  Obs.  4.) 

II.  Then  add  or  subtract  as  in  Simp)le  Xwnbers.  (Sects.  II.  and 
III.,  General  Rules ) 

III,  Place  the  separating  point  in  the  result  directly  under  the 
other  points.  ' 

Note. — lu  subtraction,  of  course  we  place  the  greater  number  at  the  top. 

Proof. — The  same  as  in  Simple  Numbers.  (Sect.  II.  Art.  2. 
Obs.  6.  and  Sect.  HI.  Art.  2.  Obs.  7.) 

Whece  do  the  separating-  points  fall  in  these  examples?  Which  number  do 
we  ni'ak^  the  minuend  in  subtracting?  Why?  What  is  the  rule  for  adding 
or  subtracting  Decunals  or  Federal  Money?     What  is  the  proof? 


138  COMMON  ARmiMETic.  Sect.  VIII . 

t&J^  EXERCISES    FOR    THE    SLATE. 

1.  A.  lias  $I8,B.  hasSl4.06,  C  has  $20.^^75,  D  has  1 6. 005, 
and  E.  has  $12.3125.     How  much  have  they  alll 

Ans.  $70,705. 

2.  How  much  has  A .  more  than  B.  in  the  last  example? 

Ans.  $3.94. 

3.  In  the  same  example,  how  much  has  A,  B.  and  C.  more  hjm 
D  and  E.]  Ans.  34. OT. 

4.  A  faiTner  has  grain  as  follows  :  of  wheat  123^0993  bushels  ; 
of  oats,  109.6  bushels-  of  corn.  03.97  bushels;  of  rye,  75.004 
bushels;  and  of  barley,  58.3267  bushels.     How  much,  grain  has  he? 

Ans.  450  bushels. 

5.  How  much  more  \'y  he  at  has  he  than  corn? 

Ans.  39. 1293  bushels. 
G%.  How  much  more  wheat  and  oats  has  he  than  all  other  kinds- 
grain?  Ans.   15.398  bushels. 

7.  A  gentleman  bought  a  eoat  for  $15.75  ;  a  hnt  for  $5. 375  ;  a 
pair  of  boots  for  $5.06  ;  and  an  overcoat  for  23. 875;  Mow  much 
did  he  pay  for  the  whole?  Ans.  ^B50.06. 

8.  How  Biuch  did  the  eoat  cost  moye  than  the  hat? 

Ans.  $10.3751 

9.  How  much  did  the  coat,.  ]mt,  and  boots  cost,  more  than  ^fie 
overcoat?  Ans.  $2.31. 

10.  A  merchant  sold  several  pieces  of  cloth»  as  follows  :  the  first 
piece  contained  16.8  yards;  the  second,  19.006  yards  ;  the  third, 
22.14  yards  ;  and  the  fourth,  22.054  yards.  How  much  was  there 
in  all?'  '  Ans.  80  yards. 

1 1.  How  much  did  the  thiyd  piece  contain  more  than  the  fourih? 

Ans.   .086  yards. 

12.  Ho. V  much  did  the  second  and  third  pieces  -contarn,.  o.ore 
than  the  tlrst  and  fourtfr?  Ans.  2.292  ytirds. 

13.  A  merchant  ha^  due  to  him  the  following  sums  :  J27  dol- 
lars, 7  cents  ;  96  dollars,  6\  cents  ;  47  dollars,  5  mills  ;  57  dolhiFs, 
8-^-  cents  ;  19  dollars,  18j  cents  ;  and  25  dollars.  How  much  has 
he  owing  him  in  all?  Ans.  $371 .41. 

14.  From  $10  take  ^j  a  esnt?  Ans.   9.995. 
^5,  From  SI  take  2  thousaidths.                            Ans.   80.998. 

16.  Add  toge  her  10  units,  3  hunlredths  ;  5  ten  thousand ihs  : 
123456  millionths  ;  762  units,  ,397844  milliorrlhs,  and  12  units, 
8972  ten  thousandths.  Ans.  785.449. 

17.  From  18  units  take  4  billionths.  Ans.   17999999996. 

18.  From  18  dollai-s,  75  cents,  take  88 J  cents. 

Ans.  $17.86^. 

19.  From  5  dollars  take  3  mills.  Ans.  $4,997. 

20.  Ad  J    together  5  eagU's,  .Cj  dimes;  6  eaoles,   3  dollars;   12 


Art.    11.    MULTIPUCATIOX    OF    DECIMALS  OR    FEDERAL    MONEY.    189 

cents  ;  2  dollars,    4  mills  ;  218  mills;  57689  mills,  and  15  eagles. 

Ans.  ^323.531. 
21.  From  9  eagles  take  9  dime?.  Ans.  ^89. 10. 

Article   II.     Multiplication  of  Decimals,  or  Federal  Money. 
Ex.  1.  Multiply  9.25  by  6. 
Oj^erat'on. 

9.25  In  this   example,  we   multiply   as    usual,  and 

6         poin     Oif  two   decimals  in   the  product,    because 

there  are  two  decimals  in  the  multiplicand.     The 

Ans.  55.53         correctness  of  this  example  may  be  sliown  by  the 
following 

lUustrct'on  — 9 . 25  =  9^~,%  =  ^%  ;    jU  X  Q  =  tVo"  =  55fVV 
=  55.50. 

2.  A  gentleman  bought   12  yards  of  cloth,    at  ^2.375  a  yard. 
How  much  was  the  cost  of  the  whole? 

Oj^erafion. 

2.375  In  this  exrmple,  we  multiply  and  point  off  ex- 

12  actly  as  in  the  tirst  example.     The  correctness  of 

^ this  operation  may  be  shown  in  the  same  manner 

Ans..S28.500  as  in  the  last. 

3.  Multiply  7.42  by  8.35. 

Operation. 

In  this  ex  rnple,   we  point  off  four  d  ci- 

mals  in  the  prodi  ct,  because  there  are  two 

decimals  in  ihe  multiplicand,  and  two  in  the 

multiplier,  making  4  in  both. 

We  prove  this  to  be  correct,  thus  :  7.42  = 

7  -2    7  :-j  .   p   or «_?5_ l'±L  •    li?   V 

'  TTTo  —  To'o"  »    "  •  *^*    —  ^10  0  —  10  0'     1  0  u    /^ 

Too   —     1  0  tTTT  J     —  "  M  0  0  0  0  u  1  .  ^  c;  /  *-/,    ti.^ 

Ans.  G  1.9570         before. 

4.  Multiply  .04  by  .07. 

Oj)erafion. 

.04  In  this  exam,  le,  there  are  but  two  figures  in  the 

.07  product,  therefore   we  prefix   ciphers  to  obtain  tlie 

requisite  number  of  dtcimnls.     The  correctne  s  of 

Ans.  .0028  this  operation  may  be  shown  thus  :  .04  =  yjy  ; 
.07  =  jU  ;  (Art.  1.  Obs.  2.)  yjo  X  tfo  =  t/-,W=  -0028,  as 
before.  (Art.  1.  Obs.  3.) 

Tlif^  illustration  shows  why  the  ( iphers  should  be  prefixed,  rather;, 
than  annexed,  to  the  product.  .1 


7. 

42 

8 

35 

37 

10 

222 

G 

5936 

140  COMMON    ARlTIIMETiO.  Scct.  VIII. 

From  these  examples  and  illustrations,  we  derive  the  following 

RULE  FOR  THE  MULTIPLICATION  OF  DECIMALS,  OR  FEDERAL 

MONEY 

I.  Wrl/e  the  mwibers,  and  midtijjly  as  in  Simple  Xuynhers. 
(Seet,  iy„  General  Rule.) 

II.  Point  off  as  many  decimals  in  the  irrodud,  as  there  are  dec- 
imals in  both  the  multiplicand  and  mtdtiplier. 

III.  If  there  are  not  as  many  decimals  in  the  product  as  in 
both  factors,  prefix  ciphers  to  it  until  the  requisite  number  of  deci- 
mals are  obtained. 

Proof. — The  same  as  in  Sinir)le  JVumbers.  (Sect.  TV.  Art.  2. 
Obs.  7.) 

1.  When  the  multiplier  is  a  composite  number,  we  may  often 
contract  the  operation,  the  same  as  in  Simple  Numbers. 

2.  To  multiply  by  10,  100,  1000,  &c.,  we  need  only  remove  the 
separatrix  1,  2,  3,  d'c,  places  to  the  right,  annexing  ciphers  if 
necessary. 

Thus  :   1 .6  X  1000  =  1600.0,  or  simply  1600, 

EXERCISES    FOR    THE    SLATE. 

1.  How  much  would  14.5  yards  of  cloth  cost,  at  $1,375  per 
yard?  Ans.  ^19.9375. 

2.  A  fiirmerhas  12  bags,  each  containing  2.75  bushels  of  wheat. 
How  much  has  he  in  all?  Ans,  33  bushels 

3.  If  he  sells  it  at  $1 .  173  per  bushel,  how  much  will  it  come  to? 

Ans.  $38,709. 

4.  How  much  would  112  pounds  of  coffee  cost,  at  0.117  per 
pound?  Ans.  S 13. 104. 

5.  How  much  would  216  pounds  of  sugar  cost,  at  ^0.071  per 
pound?  Ans.  815.336. 

6.  How  much  grain  in  \Qi  boxes,  each  containing  77.93  bushels? 

Ans.   1246.  88  bushels. 

7.  How  many  yards  of  cloth  in  27  pieces,  each  containing  32.78 
yards?  Ans.   885.06. 

8.  How  much  would  1000  pounds  of  sugar  cost,  at  $0.07  per 
pound?  Ans.  f  70. 

9.  Multiply  .0763  bj  2.16.  Ans.    .164808. 

10.  Multiply  2.97  by  .C042.  Ans.    .012474. 

11.  Multiply  12.62  by  81  J.  Ans.   1031. 68^2- 

12.  Multiply  .276  by  .C0J437.  Ans.   .000120612. 

What  is  the  rule  for  Muitipiicalion  of  Decimals  or  Federal  Money?  Ex- 
plain, by  Ex.  1,  2,  3,  and  4,  why  this  rule  is  correct.  How  do  we  prove  the 
operation?  When  the  iniilti|)lier  is  a  composite  number,  how  may  we  proceed? 
How  may  we  multiply  by  10,  100,  1000,  &c.^ 


Art.   II.    MULTIPLICATION  OF  DECIMALS  OR  FEDERAL  MONEY.        141 

13.  How  much  would  97  pmmds  of  pork  cost,  at  $3.50  per  hun- 
dred? '  Ans.  $3,395, 
Operation. 
3.50             The  result  obtained  is  evidently  the  cost  of  9700 
97         pounds,  (because  the  price  is  so  much  per  hundred,) 

whereas  we  only  wish  the  cost  of  97  pounds.     Now 

2450         97  is  y^-o  of  9700,  therefore  the  cost  of  97  pounds  is 
3150  but  T~  the  cost  of  9700  pounds.     Then  $339.50 

100  =  $3.3950.  It  will  be  perceived  that  the  lat- 


23 

500 

528 

75 

4112 

5 

$339.50  ter  result  is  obtained  by  removing   the  separating 

point  two  additional  places  to  the  left. 
The  correctness  of  this  last  point  may   be  shown  thus  :  339-50 
=  339yVa  =  'rir-    'tIV  ^  100  =  IJiJJ  (Art.  6.    Obs.  2.  a.) 
iil|2.  ^  3yWoV  =  3-3950,  as  before. 

14.  How  much  would  794  feet  of  boards  cost,  at  $5,875  per 
thousand?  Ans.  4.664+. 

Operation. 
$5,875  The  result  obtained  is  evidently  the  cost  of 

794  794000  feet,  (as  the  price  is  so  much  per  thou- 
sand,) whereas  we  only  wish  the  cost  of  794 
feet.  Now  79.4  feet  is  y^Vo  oi  794000,  and 
therefore  the  cost  of  794  feet  is  yoV^  o^  the  cost 
of  794000.  Then  $4664.750  -~  1000  = 
$4.664750.  It  will  be  perceived  that  the  latter 
$4664.750         result  is  obtained  by  removing  the  separating 

point  three  additional  places  to  the  left. 
llhMtralion.—4QQ'\.15Q  =  4664yV^V  =  '  WoV"  ;    ' VoV/"  "v- 
1000=.i5J±l|^   (Art.  6.  Obs.  2.    a.)     1^ - i||  =  4yVo'oWo  = 
4.664750,  as  before. 

Remark— The  sign  of  Addition  placed  at  the  right  of  an  answer,  signifies  that 
the  answer  is  not  complete,  or,  that  there  iaa  remainder- 
Hence — When  buying  or  selling  articles  by  the  100  : 
Obs.  1.     Point  off  tw)  additional  places  at  the  rigid  of  tlie  product. 
When  buying  or  selling  articles  by  the  1000  : 
Point  of  three  additional  places  at  the  rigid  of  the  product. 

Remark  1.— C.  stands  for  100,  and  M.  stands  for  1000;  from  two  Latin 
words,  Cen^M7nand  'Mxile,  which  signify  hundred  and  thousand. 

2. — The  learner  will  remember  that  in  all  cases  of  multiplication,  whether  of 
Simple,  Compound,  or  Decimal  numbers,  the. multiplier  is  to  be  considered  an 
abstract  number.    (Sect.  IV.  Art.  2.  Obs.  6.  Rem.  1.) 

How  is  the  final  result  obtained  in  Exs.  13  and  14?  Show  why  this  is  cor- 
rect? What  does  the  sign  of  addition,  placed  after  an  answer,  signify?  How 
do  we  ^jfoceed  in  buying  and  selling  arlicles  by  the  100?  By  the  lOOO?  For 
what  docs  C.  stand?  M.?  From  what?  What  must  be  observed  in  all  cases 
of  Multiplibalion,  whisthet  of  Simple,  Compound,  or  Decimal  Numbers? 


142  COMMON    ARlTIIMETir.  ScCt.     Vill. 

15.  How  mucli  would  18275  brick 'cost  ,  at  •'$4. 37  V  perM  ? 

Ans.  |79.953-f-. 

16.  How  miifli  would  276  bunches  of  slnugles  cost,  each  biincli 
containing-  1242,  at  .^-3.12^  per  M.?  Ans.  ^1071.225. 

17.  Ho\v^  mik'h   would   4723  pounds  of  beef  cost,  at  $3.75  per 
C?  '  $Ans.   177.1125. 

18.  How  much  would  1623  feet  of  lumber  <  ost,  at  ^4.62!r  loer 
C?  Ans.  ^750.683+. 

19.  How  much  would  5273  pounds  of  pork  cost,   at  t'^4.25  per 
C?  Ans.  $224.1025. 

20.  How  much  would  2314  feet  of  mahoo-anj  cost,   at  ^61.25 
perM.?  "Ans-  $148.6745. 

Article  12.     Division  of  Decimals,  or  Febeeal  Money. 

Ex.  1.  Divide  .84  by  .7.  : 

Ojjeraiion.Vid^ 

Obs.  1.     As  we  have  said  before,  the  dividend  .7). 84 

in  Division   answers  to  the  product  in  iVlultiplica-  

tion  ;  (Sect.  V.  Art.  2.  Obs.  7.)  and  as  we  point  Ans.  1.2 
off  as  many  decimals  in  the  product  as  there  are  decimals  in  both 
factors,  it  follows  that  the  quotient  in  Divifiiori  of  Decimals  must  al- 
ways contain  as  many  decimal  places,  as  the  decimals  i/n  the  dividend 
exceed  those  in  the  divisor  ;  the  divisor  and  quotient  being  the  factors 
which,  multiplied  together,  produce  the  dividend.  Thus,  in  the 
last  example,  the  dividend  has  two  decimal  places,  and  the  divisor 
but  one ;  therefore  we  point  off  one  decimal  place  in  the  quotient. 

This  point  may  also  be  illustrated  as  follows  : 

•  ^  ^  —  10  0'      •  '   —  1"  0"  '     Too     •     10  —  Too     ^      f    —  10  *  To   — 

1 . 2,  as  before.   (Art.  6.   Obs.  7.)  quotient.   1.2 

The  example  may  be  proved  thus  :  divisor-       .7 

divid'nd   .84 
2.  Divide  .  1476  by  1.8.  Ans.    ,082. 

Operation. 

We  perform  this  operaiion  by  Long  Division.  1 .  8).  1476).082 

As  the  quotient  does  not  contain  as  many  signifi-  144 

cant  figures,  as  the  decimals  in  the  dividend  ex-  ■ 

ceed  those  in  the  divisor,  we  supply  the  deficiency  36  •»'* 

by  prefixing  ciphers.  36  ^.^ 

Analytic  Illustration.—  .  1 476  =  t/oVo  ;   1  •  8  =  --  ''"' ' 

-18      18.      14  7fi         .18  tT7R\ytO  n2  (\(\ 

■'To'  —  T"5"»  T'5""oo-5' "^  To"  =  TiTo"  (To"  A  ys  =  T"o""o""?)r  =  ^JvJ 

.082.  (Art.   7.  Obs.  3.)     This  illustration  shows 

why  v,'o  prefix,  rather  than  annex  ciphers  to  the  quotient. 

How  many  decimal  pUet^'S  should  the  quotient,  ia  Pivisioii  of  Peciniais 
contUin?    Wby^dt  >  -^. 


Art.    12.         DIVISION  OF  DECIMALS  OR  FEDERAL  MOENY.  143 

3.  Divide  2i)70  by  .12. 

Ojycraiion. 
in  tills  example,  we  annex  two  ciphers  to  the         .12)2970.00 

liividend  (o  make  the  number  of  decimals  equal  — 

to  those  in  the  divisor ;  and  on  account  of  this       Ans.     24750 
equaUt}',  o  ir  quotient  is  a  whole  number. 

lllusf ration.— .  1 2  ==  yV, ;  2970  H-  t  Jo  =  ^  V  '  X  VV  =  ''  '  W  "  > 
=  24750,  as  before. 

4.  A  man  bought^)  pair  of  boots  for  $42.75.  How  much  was 
that  apiece? 

Operation. 
We  proceed  in  Federal   Money,  in  all  cases,         9)^42.75 

exactly   as   in   Decimal    Fractions.      (Art.    10.  

Obs.  2.)     Hence—  ^4.75 

To  perform  operations  in  Division  of  Decimals,  cr  Federal 
Money,  we  have  this 

Rule. — I.  Write  Hie  mmibers,  and  divide  as  in  Sim2)le  lumbers. 
(Sect.  V.   General  Rule.) 

II.  Poi?ii  f^'as  many  decimal  places  in  the  quotient,  as  the  number 
of  decimals  in  the  dividend  exceed  those  tn  the  divisor.  (Obs.  1.) 

III.  ^  there  is  not  a  sufficient  ninnher  of  signifxantjigiires  in  the 
quotient  to  poird  off  for  decimcds,  supyply  the  deficiency  by  prefixing  ci- 
phers. (Ex,  2.) 

Proof. — The  same  as  in  Simple  Ximihers. 

Remark  1. — When  the  nnmber  of  the  decimals  in  the  dividend  is  equal  to 
those  in  Xhc  divisor,  the  result  is  a  whok  number.  (Ex.  .3.) 

2. —  hen  th^  number  of  decimal  places  in  the  dividend  is  not  equal  to 
those  in  the  divisor,  annr-x  cipher.'^  to  the  riorht  of  the  dividend  until  they  are 
equal.  (Ex.  3  )     The  result  in  this  case  is  also  a  whole  number. 

Note. — Tn  both  these  cases,  the  divisor  is  supposed  to  be  less  than  the  divi- 
dend, and  also  that  there  is  no  remainder. 

3. — If  tliere  is  a  remainder  after  the  division  has  been  performed,  ciphers  may 
bennnexpd  to  the  remainder,  and  the  division  continued  ;  but  nfter  the  number 
of  deciirml  places  in  the  dividend  are  equal  to  those  in  the  divisor,  each  cipher 
annexed  produces  an  additional  decimal  in  the  quotient. 

This  is  evident  fron>  Art.  1,  Obs.  7  and  9.  For  1 .4  =  1  .40  =  1  .400  = 
1.4000,  &c.,  and  thus  we  may  increase  the  decimals  in  the  dividend  with- 
out allering  the  decimals  in  the  divisor,  and  in  this  way  increase  the  deci- 
mal.'j  in  the  quotient. 

What  is  the  rule  for  Division  of  Decimal.*!,  or  Federal  Money?— 
What  is  tliH  m-^ihoH  i  f  Proof?  When  the  number  of  decimals  in  the  divi- 
dend are  equal  to  tl  ose  n  the  divisor,  what  is;  the  result?  When  the  num- 
ber of  decimal*  in  the  ''ividend  is  not  equal  to  those  in  the  divisc,  how  do 
wp  pro'^'^ed?  What  is  the  result  in  this  c;ise?  If  there  is  a  remninder  after 
t'lP  divt^V,n  hap  been  perf.  v-\f'd,  ho\v  aviy  we  proceed?  What  must  be  ob- 
served in  this  cage?    ExpiuiJi  why  this  la  correct. 


144  COMMON    ARITHMETIC.  Scct.  VJIL 

Note. — In  common  business  calculations,  it  will  be  sujfficiently  exact  to 
extend  the  quotient  to  three  or  four  places  of  decimals  ;  but  when  great 
accuracy  is  required,  it  should  be  extended  farther. 

4. — When  the  divisor  is  a  composite  number,  we  may  often  shorten  the 
operation,  as  in  whole  numbers.  (Sect.  V.  Art.  4.  Obs.  3.) 

5. — To  divide  by  10,  100,  1000,  &e.,  we  need  only  remove  the  separating  point 
as  many  places  to  the  left,  as  there  are  ciphers  in  the  divisor;  and  if  there  is 
not  a  suffcienctj  of  figures,  supply  the  deficiency  by  prefixing  ciphers. 

Illustration.— DW\&Q  1  by  1000.  .  1  =  yV  ;  yV -^-  1^00  =  tuWtt 
(Art.  6.  Obs.  2.  a.)  yoUo  =  -OOOl-  (Art.  7.  .Obs.  3.) 

EXERCISES    FOR    THE    SLATE. 

1.  If  6  pounds  of  tea  cost  ^4.50,  how  much  is  that  per  pound? 

Ans.  $0.75. 

2.  If  1 2  pounds  of  coffee  cost  $  1 .  50,  how  much  is  that  per  pound? 

Ans.  J$0.12|. 

3.  If  18  yards  of  cloth  cost  $27,  how  much  is  that  per  yard? 

Ans.  $1.50. 

4.  If  26  acres  of  land  cost  8232.44,  how  much  is  that  per  acre? 

Ans.  ^8.94. 

5.  If  150  pounds  of  but  er  cost  $24,  how  much  is  that  per 
pound?  Ans.  $0.16. 

6.  If  124  bushels  of  wheat  cost  $139.50,  how  much  is  that  per 
bushel?  Ans.  $1,125. 

7.  If  464  pounds  of  feathers  cost  $116,  how  much  is  that  per 
pound?  Ans.  $0.25. 

8.  If  24  chairs  cost  $60,  how  much  is  that  apiece? 

Ans.  $2.50. 

9.  If  it  cost  a  man  $2,375  a  week  for  board,  how  long  will  $228 
last  him?  Ans.  96  weeks. 

10.  A  mechanic  received  $144  for  doing  a  piece  of  work,  which 
took  him  64  days.     How  much  did  he  receive  per  day? 

Ans.  $2.25. 

11.  Divide  45  by  .15.  Ans.  300. 

12.  Divide  2.88  by  1.2.  Ans.    2.4. 

13.  Divide  .20736  by  288.  Ans.      .00072. 

14.  Divide  13.64589  by  2.19.  Ans.  6.231. 

15.  Divide  1329.6  by  .24.  Ans.  5540. 

16.  Divide  1.4112  by  21.  Ans.   .0672. 

17.  Divide  1  tenth  by  10.  Ans.   .01. 

18.  Divide  10  by  1  tenth.  Ans.  100. 

To  how  many  places  of  decimals  should  the  quotient  be  extended  in  common 
business  calculations?  When  should  it  be  extended  farther?  How  may  we 
proceed  when  the  divisor  is  a  composite  number?  How  may  wo  divide  by  10, 
100,  idOO,  &c.?     Explain  why  this  is  corre&t? 


Art.    12.       DIVISION    OF    DECIMALS    OR    FEDERAL   MONEY.  145 

•     19.  Divide  1728  billionths  hy  288  thousandths. 

Ans.   .000006. 
20.  Divide  221  thousandths  by  17  billionths. 

Ans.   13000000. 

MISCELLANEOUS    EXERCISES    FOR    THE    SLATE  I 

Involving  the  princ'qjles  of  Decimal  Fractions  and  Federal  Money 

1.  A  man  has  owing  to  him  as  follows  :  from  A.,  f  17;  from  B., 
86. 12| ;  from  C,  ^27.06|  ;  and  from  D„  $0.81^.  How  mich 
has  he  owing  to  him  in  all?  Ans.  $51. 

2.  How  much  does  C.  owe  him  more  than  A.  and  B.? 

Ans.  83.93|. 

3.  How  much  does  C.  owe  him  more  than  all  the  rest 

Ans,  $3.12^, 

4.  A  gentleman  having  $100,  spent  $25  for  clothin-,  $15  for 
books,  $6.62^  for  riding  in  the  stage,  and  $40.87^  for  jewelry  ; 
how  much  had  he  left?  Ans.  $12.50. 

5.  How  much  more  did  he  give  for  his  jewelry  than  for  his  cloth- 
ing? Ans.  $15.87|. 

6.  How  much  more  did  he  spend  for  clothing  than  for  riding  in 
the  stage?  _  Ans.  $18.37|. 

7.  A  farmer  received  for  his  marketing  $27,125;  of  this,  he 
spent  $2,375  for  sugar,  $1. 1875  for  coffee,  $2.25  for  tea,  $0.75 
for  spice,  $14.0625  for  cloth,  and  took  the  rest  home.  How  much 
did  he  take  home  ?  Ans.  $6 .  50. 

8.  How  much  more  did  he  spend  for  cloth  than  for  all  the  other 
things?  Ans.  $7.50. 

9.  Add  together  fourteen  units,  six-tenths  ;  two-hundredths ;  nine 
units,  forty-seven  thousandths  ;  seventy-six  ten-thousandths  ;  and 
one  unit,  two  ten-thousandths.  Ans.  24.6748. 

10.  Add  together  7  units,  602  thousandths;  18  units;  9  hun- 
dredths ;  43  units,  26  hundredths  ;  7  units,  8071  ten-thousandths; 
and  286  units,  4  tenths.  Ans.  363. 1591. 

11.  From  18  units,  take  63  thousandths.  Ans.   17.937. 

12.  From  1  unit  take  3  millionths.  Ans.   .999997. 

13.  How  much  would  16  sacks  of  coffee,  each  containing  173 
pounds,  cost,  at  7-|  cents  per  pound?  Ans.  $207.60. 

14.  How  many  bushels  oi  wheat,  at  $0:875  per  bushel,  would 
it  take  to  pay  for  the  coffee,  in  the  last  example? 

Ans.  237/j. 

15.  A  merchant  bought  46  bags  of  oats,  each  containing  2.5 
bushels,  at  $0. 1875  per  bushel.     How  much  did  they  cost  him? 

Ans.  $21.5625. 

16.  'Ba  paid  for  them  with  coffee,  at  10  cents  per  pound.  How 
many  pounds  did  it  take?  Ans.  216.626. 

8 


146  COMMON    ARITHMETIC.  ScCt.    VIII. 

17.  How  much  would    47398  feet   of  lumber  cost,    at  $3.50 
per  C?  Ans.  $1658.93. 

18.  How  much  would  8372  pounds  of  pork  cost,  at  $3. 875  per 
C?  Ans.  $324,415. 

19.  How  much  would  2146  feet  of  boards  cost,  at  $10  per  M.? 

Ans.  $21.46. 

20.  How  much  would  1623  shingles  cost,  at  $6. 125  per  M.? 

Ans.  $9.94-1-. 

21.  If  a  man  earn  $0,625  per  day,  how  much  can  he  earn  in  a 
year,  (365 days.)  Ans.  $228,125. 

22.  If  he  spends  $0,375  per  day,  how  much  will  he  have  left  at 
the  end  of  the  year?  Ans.  $91 .25. 

23.  How  many  yards  of  cloth  can  he  buy  with  this  money,  at 
$3,625  per  yard?  Ans.  $25/^'. 

24.  Multiply  62.7  by  100.  Ans.  6270, 

25.  Divide  8.726  by  100.  Ans.   .08726. 
2^,                   Just  fifteen  pair  of  ladies*  gloves. 

For  forty  dimes  had  I : 
How  many  pair  of  that  same  kind 
Will  forty  eagles  buy  ?  Ans.   1090. 

Article  13.     Bills,  Accounts,  &c. 

Obs.  1 .  A  Bill,  in  dealings  with  merchants,  and  others,  is  a 
written  paper,  containing  a  statement  of  the  particulars,  and  total 
cost  of  the  goods  sold. 

To  find  the  total  cost  of  a  Bill : 

Obs.  2.  Find  the  cost  of  each  particular  at  the  price  mentioned, 
and  the  sum  of  these  vMl  l>&  the  cost  required. 

Required — the  cost  of  the  several  articles,  and  the  total  sum  in 
following  bills  : 

(1) 

Columbus,  Jan.  6th,  1847. 
Pet^r  Paywell, 

Bought  of  James  Freeman,  c&   Co. 

16  pounds  Coffee, at   $0. 12^  per  pound, 

18       "       Sugar, ,  at        .09         " 

12       "       Tea, at        .87i^       *• 

26       **       Saleratus,. at        .04        ** 

25       '*       Raisins,.. at       .16|       " 

Total  cost, $19.32|; 

'wi«t  jirliJ?  lJ<?w  do  we  find  tho  total  coat  of  ^  fell? 


Art.  13.  BILLS,    ACCOUNTS,  &iC  147 

(2-) 

CLEyjEJLAND,  Feb.  2d.,  1848. 
^S',  J".  Arsdn, 

Eeug^t  of  Charles  Mavtin. 

6     yards  Broadclotli,  . , at  $4.62^  per  yard 

4        ''  Cambric,. at      .  12|       " 

2^  dozen  Buttons, at       .25    per  dozen 

9     skeins  Silk, at      -06|^  per  skeia 

4^  yards  Wadding, . . . .  ^  ^ at      -  09     per  yard 

Totalcost, ^29.84J. 

(3.) 

OiNdNNATi,  June  7th,  1848- 
Henry  Plylmrd^ 

To  LevAs  Anderson,  <^  Co,     Dr, 

For  4  copies  Davies'  Bourdon, .... .  at  8K  37|-  eacli 

^*  3     **       Anthon's  Caesar, at     1 .  00       " 

^'4     ^*       Leverett's  Dictionary, at     4.87^     « 

"'  5    ^       Gredt  Testament,,.  . ... at     i,12|-    ** 


Total  cost, ^38-  1 2^^. 

(4.) 

Delaware,  March  1st,  1847. 

To  Auyustvj%  Rdcharts,  <fr  Cb.     Dr. 

For    897  feet  Boards, at  80. 87^  per  C. 

"    1247    **     Plank, at      1.12| 

^*      479    "     Scantling, at        .75  " 

^i    2479   **     Flooring, at      1.25  « 

*'    8762    ''     Shingles, at      4.37-^perM, 

Total  cost, ....  $94. 788+, 

Received  Payment. 
^, '  Au^stus  Rficharts,  xk  Co^ 


148'  COMMON    ARITHMETIC.  Sect.    IX 

(5.) 

Pittsburg,  May  1st,  1848. 
Thomas  Thrifty, 

Sold   Wm.  Trader,  c^  Co, 

1^6 bushels  Wheat, _ at  $K  12^  per  bushel 

1423       "       Barley, at       .56^ 

4679       "       Corn, at       .22 

3716       "       Oats, at       .25 

2742  pounds  Cheese, at       .08    per  pound 

In  return,  he  received — 

In  Cash, 85000.00 

144  yards    Satinet, at  $1 .  25  per  yard 

68     *'        Silk, at     1.06^     " 

176     *'         Muslin, at       .111     '* 

168     ''         Calico, at       .16|-     '* 

1236  pounds  Coffee, at       .  1  If  per  pound 

1374       "       Sugar, at       .08|     " 

Required — the  difference  between  the  accounts,  and  in  whose 
favor.  '■  Ans.  $308  .  38|-  in  favor  of  Thomas  Thrifty. 


SECTION  IX. 
COMPOUXD  NUMBERS. 

Article  I.     Definitions,  &c. 

Obs.  I.  When  \he,  ratio  of  increase  is  the  same,  numbers  are 
called  Simple.  Thus:  156,28,  5  dollars;  13  yards;  147  miles, 
&c.,  are  called  simple  numbers,  because  each  order  has  ten  times 
the  value  of  the  next  lower  order. 

Obs.  2.  When  the  ratio  of  increase  varies  in  the  different  orders 
or  numbers,  they  are  called  Compound.  Thus,  12  bushels,  3  pecks, 
6  quarts  ;  3  miles,  30  rods,  14  feet,  t&o.,  are  called  compound  num- 
bers, because  in  some  of  these  it  takes  more,  and  in  some  less,  than 
ten  units  of  one  order  to  make  one  unit  of  the  next  higher  order. 

Remark.—  Compound  numbers,  by  somo  authors,  are  called  Denominate 
Numbers. 

Obs.  3.     The  only  difference  between  Simple  and   Compound 

When  are  nambers  called  Simple  numbers?  Give  examples.  When  aro 
they  called  Compound  numbeasl    Give  examples. 


Art.  2,  REDUCTION,  149 

numbers,  is  this  :  In  Simple  numbers,  figures  increase  umfonrdy  hy 
10  ;  that  is,  it  takes  10  units  of  each  order  to  make  1  unit  of  the 
next  higher  order  ;  but  in  Compound  numbers  figures  increase  differ- 
ently, sometimes  taking  mor-e,  cmd  sometimes  less,  tJum  10  units  of  one 
order,  to  make  1  unit  of  the  next  Jiigher  order. 

In  the  former  one  table  alone  is  necessary  ;  in  the  latter  many  ta- 
bles are  required- 

Obs.  4.  The  Tables  in  Co^mpound  numbers  ieacli  how  many  units 
it  takes  of  one  order  to  make  a  unit  of  the  next  higJier  order.  The 
Numejyition  Table  fceadies  the  same  in  Simple  numbers. 

Remark. — The  differeat  orders  in  Compoand  u«mbej'sare  called  denomina- 
tions. 

Obs.  5.  In  Simple  nwvhers,  the  units  of  any  order  can  always 
be  expressed  by  one  figure  ;  in  Compound  numbers,  it  sometimes 
takes  two,  thi'ee,  or  even  four  figures  to  express  the  units  of  a  sin- 
gle order. 

Likewise,  tliKS  sum,  or  difference  of  any  order,  in  Simple  numbers, 
^an  always  be  expressed  by  one  figure ;  but  in  Compound  numbers, 
the  sum,  or  difference,  can  sometimes  be  expressed  by  one  figure, 
and  sometimes  it  requires  more  than  one  figure  to  express  it. 

Remark. — The  learner  will  bear  in  mind,  that  we  carry  1  to  the  next  higher 
order,  as  often  as  we  ebtaln  a  sufficient  number  of  units  of  the  lower  order,  to 
tnake  l  unit  of  this  higfier  order,  whether  in  Simple  or  Compound  numbers 
Hence — 

Obs.  6.  TJie  pri?icip!es  of  Simple  and  Compound  iiumbers  aretlie 
^ame. 

Article  2,     Reduction. 

Obs.  1 .  Changing  numbers  from  one  denomination  or  kind  to  <m- 
other,  without  altering  their  value,  is  called  Reduction.  Thus,  there 
are  4  pecks  in  a  bushel ;  ihen  in  2  bushels  there  are  8  pecks  ;  there- 
fore, 8  pecks,  or  2  bushels,  express  the  same  quantity. 

Remark  — Reduction  is  of  tw©  kinds — Descending  and  Ascending. 

What  js  the  difference  between  Simple  and  Compound  numbers?  How 
many  tables  are  necessary  in  th«  former?  Are  more  tables  than  this  required 
in  the  latter?  What  do  the  tables  in  Compound  numbers  teach?  What  does 
it  leach  in  Simple  numbers?  Wiiat  are  tiie  differs ni  orders  in  Compound  num- 
bers ciilied?  How  many  figures  d«es  it  take  to  express  a  unit  o(  a  eingle  ord^r 
in  Simple  numbers?  iHow  mauy  in  Compound  numbers?  How  many  fig- 
ures does  it  take  to  express  the  sum,  or  difference,  of  any  order  in  Simple  num- 
bers? How  many  in  Compound  numbers?  How  often  do  we  carry  1  lo  the 
nrxt  liigher  order?  Is  this  the  same,  both  in  Simple  and  Compound  numbers? 
Whfltjjthen  is  the  difference  between  the  principles  of  the  two?  What  is  Re- 
duction? How  is  it  divided?  What  is  Reduction  descending?  What  is  Re* 
(duetifin  ascending?    ^ivean  example  illustrating  each  case. 


I&O  COMMON    ARITHMETIC,  Sect.    IX, 

Ohs.  2.  When  a  number  is  cTicmged  from  a  denomination  of  greater 
mdue  to  a  denomination  of  a  less,  value,  the  process^is  called  Reduction 
Descenbing. 

When  a  number  is  changed  from  m  denomination  of  less  value  to  Oi 
^enorrdnation  of  greater  value,  the  process  is  caMed  Reduction  As- 
cending-. 

Thus,  to  cTiange  biisli€Ts  to^^  pecks,  we  change  a  greater  de- 
noaiination  to  a  less,  and  the  process  is  called  Reduction  •Descend- 
ing. But  to  change  pecks  to  bushels,  we  change  a  less  denomina- 
tion to  a  greater,  and  the  process  is  called  Reduction  Ascending. 

There  are  4  pecks  in  a  bushel ;  how  many  peek*  in  2  bushels? 
In  3;  4;  6,   10;  12  bushels? 

How  manj  bushels  m  4  pecks?  In   8,   16;  20;  28,  3.9;  32;  44 

pecks? 

Obs.  3.     From  these  examples,  the  pupil  will  peme^we  that 
Reducthn  Descending  is  performed  lyg  Midtiplication,  and 
Reduction  Ascending  is  performed  by  cFwision.     Therefore — 
Reduction  Descending  and  Reduction  Ascending  mutually  prove 
each  other. 

RsMARS. — Compound  numbers  are  chkfly  confined  to  weights  and  measures. 

WEIGHTS. 


1.    TROY  WEIGHT. 
Obs.  4.  This  is  used  in  weighing  gold,  silver,  jewels,  liquors,  t&c. 
The  denominations  are  Grain,  ReTmyweiyht,  Ounce  and  Pound. 

TABLE. 

24  grains  (grs.) make  1  pennyweight,  marked   pwt. 

20  pennyweights **     1  ounce, *  *  oz. 

12  ounces '*    1  pound ,"  lb-. 

Rem-ark  1. — Thestandard  of  Weiglits  varies  in  different  States  in  the  Union 
In  1834  the  Gcrersiment  adopted  a  uniform  standard  for  the  use  of  the  sev-" 
era!  Custom-houses,  and  other  purp©ses.     It  is  very  de&jrable  that  this  standard 
Bhoold  be  adopted  throughout  the  Union. 

2. — The  standard  unit  of  weight  adopted  by  the  United  States  is  the  Troij 
pound  of  the  U.  S.  Mint,  which  is  tii«  same  a«  the  Imperial  Troy  poand  of 
jEngtaad,  established  by  Act  of  Parliament,  A.  I>.  I83€, 

How  is  Reduction  I>esce>nding  performed?  Reduetion  Ascending?  What 
relation  do  they  bear  to  each  other?  To  what  a;re  compound  numbers  chiefly 
applied?  For  what  is  Troy  weight  used?  What  are  the  denominations?  Re- 
peat tke  Table.     What  is  the  standard  unit  of  Weight  in  the  United  States? 


Art.  2, 


UECUCTION. 


151 


MENTAL   EXERCISES, 


1.  How  many  ounces  in  2  pounds?  4;  9;  7;  12;   13;  6;  5;  11;  8? 

2.  How  many  pennyweights  in  2  ounces?  4;  9;  6;  3;  5;  8;  10? 

3.  How  many  pounds  in  24  ounces?  36;  72,  96;  120;  132? 

EXERCISES    FOR    THE    SLATE. 


5.  Reduce  3  lbs.  7  oz,  1 5  pwts 
1 8  grains  to  grains. 

0/)ercUion, 
lbs.    oz.  pwts.    grs. 
3_._7_-_15.-.18 
12 

36 

7  oz,  added. 


43 

20 

860 
15 

875 
24 

3500 
1750 


pwts.  added. 


21000 

18  grs.  added. 


21018  grs.  Ans. 

We  first  multiply  by  12,  to  re 
duce  the  lbs.  to  oz.,  and  add  in 
the  7  oz.,  making  43  oz.  We 
multiply  this  by  20,  to  reduce  it 
to  pwts.,  and  add  in  the  1 5  pwts., 
making  875  pwts.  This  we  mul- 
tiply by  24  to  reduce  it  to  grs., 
and  add  in  the  18  grs.,  making 
2018  grs.,  as  our  answer. 

If  we  choose,  we  can  add  in 
the  numbers  given  of  the  differ- 
ent, denominations  mentally,  and 
thus'  shorten  the  operation. 

Thus:  3  X  12  =  36  ;  36  +  7 
»3  43  ;  set  down  43,  <iifc. 


6.  Reduce  21018  grs.  to  pounds 
Operaiion, 
24)21018 

2jO)87i5+18  grs.  rem. 

12)43  +  15  pwts.  rem. 

3  -1"  7  oz.  rem. 
Ans.  3  lbs.  7  oz.  15  pwts.  18  grs. 
We  first  reduce  the  21018  grs, 
to  pwts.  by  dividing  by  24  ;  the 
result  is  875  pwts.,  and  18  grs. 
remaining.  We  next  reduce  the 
875  pwts.  to  oz.,  by  dividing  by 
20;  the  result  is  43  oz.,  15  pwts. 
We  reduce  the  43  oz.  to  lbs.  by 
dividing  by  12  ;  the  result  is  3 
lbs.  7  oz.,  and  the  other  remain- 
ders make  3  lbs.,  7  oz.,  15  pwts., 
18  grs,  as  our  answer. 


152 


COMMON     ARITHMETIC. 


Sect.IX 


;5frTE. — It  would  be  a  good  plan  to  write  the  name  of  the  denominator  to 
whi<  h  each  number  belongs,  over  the  number,  (as  in  the  5th  example,)  to  pre- 
Teni  mistakes. 

From  these  examples,  we  derive  the  following 

GENERAL  RULES  FOR  REDUCTION. 


For  Reduction  Ascending : 


For  Reduction  Ascending 


I.  Divide  the  give7i  quantity  hy 
that  number  which  it  takes  of  this 

of  the 


I.  Multiply  the  highest  denomi- 
nation hy  that  number  xohich  it  takes 

of  the  'itext  less  to  make  1   of  this  denomination   to  make  one 
higher f  and  add  to  the  product  the  next  higher. 
Ttumber  given,  [if  any,)  of  the  lou)-\     II.  Proceed  in  this  manner  until 


oT  denominMtion. 
II.  Proceed  in  the  same  manner 


\ihe  whole  is  reduced  to  the  denomi- 
nation required.    The  last  quotient, 

.^7.7  •   •        7  .J       toqether  with  the  several  remainders, 

with  the  remaining  denominators,  ^v,.         s  j       -n  i    4i 

..7.7       1   1    '       J      1 4    41.     1    i'^J  any,)  annexed,  wut  be  the  an- 
untu  the  whole  is  reduced  to  the  de-}'^  r. 

•     ■ .  -7  !  swer  sought, 

nomination  required .  t^t        ^ ..    .,     „„„  j;„- •  „    .u^,.^ 

^  Note. — If,  after  any  division,  there 

Note.— If  any  denomination  is  want-;  is  no   remainder,  a   cipher   should   be 
ing,  supply  its  place  with  a  cipher,      i  written  in  the  place  of  that  denomina- 

Ition. 


7.  Reduce  3  lbs.  8  oz.  lOpwts. 
to  grains. 

9,  Reduce  6  lbs.  16  pwts.  to 
grains. 

11.  Reduce  7  lbs,  19  grs.  to 
grains. 

13.  Reduce  2  lbs.  to  penny- 
weights. 


8.  Reduce  21360  grs  to  pounds 

10.  Reduce  34968  grains  to 
pounds. 

12.  Reduce  40339  grains  to 
pounds. 

14.  Reduce  480  pennyweights 
to  pounds. 


2.    AVOIRDUPOIS  WEIGHT. 

Obs.  6.  This  is  used  in  weighing  the  articles  of  a  coarse,  heavy 
nature,  such  as  tea,  coffee,  sugar,  flour,  hogs,  grain,  c&c,  and  all  metals 
except  gold  and  silver.  It  is  also  used  in  buying  and  selling  medi- 
cines. The  denominations  are  Dram,  Ounce,  Po7nd,  Quarter, 
Hundred-weight  and  Ton. 


Why  do  we  write  the  name  of  the  denominations  over  the  numbers? 
What  is  the  rule  for  Reduction  Descending?  For  Reduction  Ascending? 
If  any  denomination  is  wanting,  how  do  we  proceed?  If  after  any  di- 
vision there  is  no  remainder,  how  do  we  proceed?  For  what  is  Avoirdupois 
weight  used?     What  are  the  denominations?     Repeat  the  Table. 


Art.  2. 


REDUCTION. 


153 


TABLE. 


16  drams  (drs.) make  1  ounce marked  oz. 


16  ounces 

26  pounds 

4  quarters 

20  hundred  weielit 


1  pound 

1  quarter 

1  hundred  weight. 
1  ton 


lb. 

qr. 

cwt. 

T. 


Remark  1. — The  Avoirdupois  Pound  of  tlie  United  States  i:3  determined 
from  the  standard  Pound  Troy,*  and  is  iu  the  ratio  of  57G0  to  7000.  That  is, 
1  pound  Troy  contains  5760  grains  ;  1  pound  Avoirdupois  contains  7000  grains 
Troy. 

2. — III  this  weight  the  words  gross  and  net  occur.  Gros$  loeigld  is  the 
weight  of  goods,  together  with  that  of  the  boxes,  casks,  or  bags  tiiat  contain 
them.     Net  weight  is  the  weight  of  the  goods  alone. 

3. — Formerly  it  was  customary  to  allow  112  pounds  for  the  hundred  weight, 
and  28  pounds  for  the  quarter  ;  but  this  practice  has  beconje  nearly,  or  quite, 
obsolete.  In  buying  and  selling  all  articles  of  commerce  estimated  by  weight, 
thelawsefmostof  the  States,  as  well  as  general  usage,  call  100  pounds  a  hundred 
weight,  and  25  pounds  a  quarter.  In  the  U.  S.  Custom-house,  and  also  in  in- 
voices of  English  goods,  and  of  coal  from  the  Pennsylvania  mines,  28  lbs.  are 
called  a  quarter,  and  2240  lbs.  a  ton. 

1.  How  many  drams  in  2  ounces?  3;  5;  4;  6? 

2.  How  many  ounces  in  2  pounds?  6;  4;  3;  5? 

3.  How  many  quarters^in  2  cwt.?  4;  9;  3;  6;  5;  7;   12? 

4.  How  many  pounds  in  32  ounces?  48;  80,  64;  76? 

5.  How  many  cwt.  in  8  quarters?  20;  27;  36;  24;  48? 


6.  Reduce  7  lbs.  9  oz.  to  drams. 
8.  Reduce  7  cwt.  3  qrs.  22  lbs. 
to  ounces. 

10.  Reduce  3  tons  to  drams. 


12.  Reduce 
lbs.  to  ounces. 

14.  Reduce  6  T 
to  drams. 


7  T.   14  cwt.   23 
lbs.  9  oz. 


7.  Reduce  1938  drs.  to  ounces. 
9.  Reduce  12752  oz.  to  hun- 


dred weight. 

11.  Reduce   1536000 
tons. 

13 
tons. 

15 
tons. 


drs.    to 


oz. 


to 


Reduce     246768 

Reduce   2564752    drs.    to 


Obs,  6.     This  is  used 
medicines. 


3.    APOTHECARIES  WEIGHT. 

apothecaries  and  physicixins  in  mixing 


How  is  the  Avoirdupoiapound  determined?  Whatis  the  ratiobetween  the  two? 
How  many  grains  in  a  pound  Troy?  In  a  pound  Avoirdupois?  WSatis  gross 
weight?     Net  weight?     For  what  is  Apothecaries  weight  used? 

a — -*:.  ^<* 

*  y\  '^onnd  Avoirdupois  is  heavier  tlian  a  poiiii<]  Troy,  but  an  ounce  Troy  is  licavicr  than 
«n    nncc  Avoirdupois. 


8a 


154  ^COMMON,  ARITHMETie.  SCCl.    lA 

The    denominations   are    Oram,    Scruple,    Dram,     Ounce,    and 
Pound. 

TABLE, 

20  grains  (grs.) mate  1  scruple, marked  sc,  or  9- 

3  scruples "      1  dram, . **       dr.  or    3. 

8  drams **      Jounce, *'       oz.  or    5. 

^;  12  ounces *'      I  pound, "  ib. 

Remark. — The  pound  and  ounce  are  the  same  in  this  weight  as  they  are  in 
Troy  weight  ;  but  the  other  denominations  are  different. 

1.  How  many  scruples  in  2  drams?  4;  6;  9;  8;  b;  7;   12? 

2.  How  many  drams  in  2  ounces?  6;  5;  7;  4;  8;   10;   12? 

3.  How  many  ounces  in  2  pounds?  4;  6;  8;  12;  5;  7;  9? 

4.  How  many  drams  in  6  scruples?  24;   18;  36;   15;  27? 
6.  How  many  ounces  in  16  drams?  32;  56;  40;  72;  96? 

6.  How  many  pounds  in  24  ounces?  72;  60;  96;  144;  108? 

7.  Reduce  1  ib.  7  3.  2  3.  1  9.|     8.  Reduce     9276     grains     to 
16  grs.  to  grains.  ipounds. 


9.  Reduce  7  ib.  6   3.  to  scru 
pies. 

11.  Reduce    6    ft).     1    9.    to 
grains. 


10.  Reduce  2034  scruples  to 
pounds. 

12.  Reduce  34580  grains  to 
pounds. 


MEASURES  OF  CAPACITY. 

1.    DRY  MEASURE. 
Obs.  7.     This  is  used  to  measure  ff rain,  fruit,  salt,  <j^c. 
The  denominations  are  Pints,  Quarts,  Pecks  and  JBtcshels, 

TABLE. 

2  pints  (pts.) make  1  quart, marked  qt, 

8  quarts **     1  peck, "      pk. 

4  pecks *'     1  bushel, **      bu. 

Remark  1. — A  quart  Dry  Measure  contains  671  cubic,  or  solid  inches. 
2 — A  Winchester  bushel  is  I8i  inches  in  diameter,  (that  is,  across  the  top,) 
and  8  inches  deep,  and  contains  2150|.  cubic  inches. 

By  this  is  meant  a  compact  bushel,  as  wheat,  oats,  salt,  shelled  corn,  &c.     A 

What  are  the  denominations?  Repeat  the  Table.  To  what  oUier  weights 
are  the  pound  and  ounce,  Apothecaries  weight,  equal?  For  what  is  Dry  mea- 
sure used?  What  are  the  denominations?  Repeat  the  the  Table,  ilow  many 
cubic  inches  in  a  quart,  dry  measure?  What  is  the  size  of  a  Winchester 
bushel?  How  many  cubic  inches  does  it  contain?  What  kind  of  a  bushel  is 
meant  by  this? 


Art.  2. 


REDUCTION. 


155 


dry  bushel,  as  apples,  peacheg,  coal,  potatoes,  coiitaing  2633  cubic,  or  solid 
inches.     The  former  is  an  even  bushel,  and  the  Utter  is  a  heaped  bushel. 

3. — The  measure  used  varies  in  different  States.  In  Connectieut,  2193  cubia 
inches  make  a  bushel.  In  New  York,  2218.192  cubic  inches  make  a  bushel. 
This  is  the  imperial  bushel  of  Great  Britain,  and  weighs  8  lbs.  Avoirdupois, 
of  distilled  water,  at  62°  Fahrenheit,  and  30  inches  of  the  barometer. 

The  Winchester  bushel  is  the  United  States  standard  unit  of  Dry  Measure, 
and  contains  77.627413  lbs.  Avoirdupois  distilled  water,  maximum  density, 
(which  is  about  41"  Fahrenheit,)  and  weighed  in  air  at  30  inches  of  the  barom- 
eter. Its  true  contents  are  2150.42  cubic  inches,  nearly,  although  2150.4  is 
usually  given. 

It  is  desirable  that  the  United  States  standard  should  be  adopted  by  the  dif- 
ferent States,  the  same  as  our  currency.  This  would  create  uniformity,  and 
thus  destroy  much  trouble  and  confusion  usually  attendant  upon  a  varia'.ion 
of  standards. 

4.  Many  persons  purchase  grain  by  weight,  instead  of  by  measure.  In  Ohio, 
and  also  in  a  t.  ajority  of  the  States,  the  weight  of  grain  is  established  by  law, 
as  follows : 


1  bushel  of  Wheat weighs 

1         "  Rye  or  Indian  Corn,       ♦* 

1         «*  Barley " 

1         «'  Oats, *• 


60  lbs.  Avoirdupois. 
56    " 

48    "  " 

33    "  ** 


Hence — To  find  the  number  of  bushels  of  grain  in  a  given  quantity  : 
a.     Divide  the  weight  of  the  grain  by  60,  56,  43,  or  33,  according  as  it  is 
Wheat,  Rye,  Corn,  Barley,  or  Oats, 

1.  How  many  pecks  in  2  bushels?     5;  8;   10;   15;   18;  23;  30? 

2.  How  many  quarts  in  2  pecks?     3;  4;  7;  5;  6;  9;  8f   11? 

3.  How  many  pints  in  2  quarts?     5;  4;  6;  3;  9;  7;  8? 

4.  How  many  bushels  in  8  pecks?     16;  32;  24;  48;  36?  ^ 

5.  How  many  pecks  in  16  quarts?     32;  64;  48;  72;  96? 

6.  How  many  quarts  in  4  pints?     12;  24;   16;  36;   18; 

7.  How  many  quarts  in  1  bushel?     2;  4;  6;  3;  5;  8;  10? 

8.  How  many  pints  in  1  peck?     2;  6;  4;  10;  5;  8;  3? 

9.  How  many  pints  in  1  bushel?     2;  4;  6;  3;  5;  8;   10? 

10.  How  many  bushels  in  32  quarts?     64;  96;   128;  160;  192? 

11.  How  many  pecks  in  16  pints?     32;  48;  80;  61;  96;  112? 

12.  How  many  bushels  in  64  pints?      128;  256;  192:  320? 

Each  of  the  following  examples  proves  the  one  opposite  : 


13.  Reduce  3  bu.  2  pks.  to 
pecks. 

15.  Reduce  6bu.  1  pk.  7  qts. 
to  pints. 

17.  Reduce  10  bu.  3  pks.  6 
qts.   1  pt.  to  pints. 


els. 

1 

els. 


14.  Reduce   14  pks.  to  bush- 
16.  Reduee  414  pts.  to  bush- 


els. 


18.  Reduce  701  pts.  to  bush - 


How  many  cubic  inches  does  the  dry,  or  heaped  bushel,  contain?    What 
j8  the'' United  States  standard  unit  of  Dry  Measure? 


156 


COMMON     ARITHMETIC. 


Sect.  IX 


19.  Reduce  37  bu.  1  pk.  5  qts. 
1  pt.  to  pints. 

21,  Reduce  68  bu.  1  pt.  to 
pints. 

28.  Reduce  121  bu.  2  qts.  to 
pints. 


els. 


20.  Reduce  2395  pts.  to  bush- 


els. 


22.  Reduce  4353  pts.  to  bush- 


els. 


24.  Reduce  7748  pts.  tob-asli- 


2.    WINE  MEASURE. 

Obs.  8.  This  is  used  for  measuring  all  liquids,  ale,  heer,  and 
milk  excepted. 

The  denominations  are  Gill,  Pint,  Quart,  Gallon,  Barrel  and 
Hogshead.  ■•, 

TABLE. 

4    gills   (gi.) make  1  pint, "  pt. 

2    pints *'     1  quart, **  qt. 

4    cuarts <•     1  gallon, "  gal. 

31|  gallons „___     "     1  barrel, "  bbl. 

63    gallons,  or  2  bbls.     *'     1  hogshead, "  hhd. 

ALSO  : 

42  gallons make  1  tierce, marked  tier. 

84  gallons,  or  2  tier.      •*     1  puncheon, **       pun. 

126  gallons,  or  2  hhd.      **     1  pipe,  or  butt, "       P. 

2  pipes '*     1  tun, *'       T. 

Remark. — The  standard  unit  of  Liquor  Measure  adopted  by  the  United 
States  is  the  wine  gallon,  containing  231  cubic  inches,  equal  to  8.339  lbs. 
Avoir,  of  distilled  water  at  the  maximum  density  ;  i.  e.,  41  ^  Fahrenheit. 

1.  How  many  gills  in  2  pints?     3;  5;  9;  7:   12;  8;   18;  4;   11? 

2.  How  many  gills  in  2  quarts?     3;  7;  9;  5;   12;   10.  6;  8;   11? 

3.  How  many  pints  in  1  gallon?     2;  5;  3;  9;  6;  5;  7;   12;   10? 

4.  How  many  gills  in  1  gallon?     2;  4;  3;  5? 

5.  How  many  pints  in  8  gills?     16;  48;  40;   12;  32;  20;  36? 

6.  How  many  quarts  in  8  gills?      16,   48;  96;  40;  20;  36;  28? 

7.  How  many  gallons  in  8  pints?     16;  48;  60;  96;  5Q;  88;  40? 

8.  How  many  gallons  in  32  gills?     96;   160;  64;   138? 

9.  Reduce    10   gal.  2   qts.  to 


gills. 

11,  Reduce  1  bl,  to  pints, 

13,  Reduce  2  hhds,  27  galls. 
3  qts.  1  pt.  3  gi.  to  gills. 


10,  Reduce  336  gills  to  gal- 
lons, 

12.  Reduce  252  p'nts  to  bar- 
rels, 

14,  Reduce  4927  gills  to  hogs- 
heads. 


For  what  is  Wine  measure  used?  What  are  th©  denominations?  Repeat 
the  Table.  What  is  the  standard  unit  of  Liquid  Measure  adopted  by  the  United 
Slates?    What  does  it  contain? 


AyT.  ^: 


REDUCTION. 


157 


15,  Reduce  IT,  1  P.  1  hhd. 
32  galls,  1  pt,  to  pints, 

17,  Reduce  7  hlids.  3  gi,  to 
gills. 


16,  Reduce  3785  pts,  to  tuns, 
1 8,  Reduce  14115  oriUs  to  hoo-s- 


heads. 


3.    ALE,   OR  BEER  MEASURE. 
Obs.  9.     This  is  used  for  measuriiug  ale,  heer,  and  milk. 

The  denominations  are   Pint,    Quart,   Gallon,  Barrel,  and  Hogs- 
head. 

TABLE. 

2  pints  (pts.) make  1  quart, marked  qt. 

4  quarts "  1  gallon, *'       gal. 

36  gallons **  1  barrel, **       bar. 

54  gallons,  or  1^  barrels, **  1  hogshead,.- **       hhd. 

Remark. — A  gallon,  beer  measure,  contains  282  cubic  inches. 

1.  How  many  pints  in  2  quarts?     4;  9;  5;  7;  6;   12;  8;   10;  9? 

2.  How  many  quarts  in  2  gallons?     3;  5;  4;  9;   12;  6;  8;  11? 
.3.  How  many  pints  in  1  gallon?     2;  4;  5;  3;  7;   12;  9;   10? 

4.  How  many  quarts  in  8  pints?     4;   16;   12;  24;   14;   18;  22? 

5.  How  many  gallons  in  8  quarts?     16;  20;   12;  48;  40;  36? 

6.  How  many  gallons  in  8  pints?     24;   16;  48;  32;  56;  40;  96? 


7.  Reduce  18  galls.  2  qts,  1 
pt.  to  pints. 

9.  Reduce  3  hhds.  3  qts,  to 
pints. 

11.  Reduce  6  bar.  1  pt.  to 
pints. 

13.  Reduce  2  hhds.  3  qts.  to 
quarts. 


8.  Reduce  149  pts.  to  gallons. 

10.  Reduce  1302  pts.  to  hogs- 
heads. 

12.  Reduce  1729  pts,  to  bar- 
rels. 

1 5.  Reduce  435  qts,  to  hogs- 
heads. 


MEASURES  OF  EXTENSION. 


Remark  1. —  The  three  kinds  of  measures  we  have  just  given  are  used  for 
measuring  liquids,  grain, fruit,  &c.,  for  which  purpose  we  hiive  different  kinds 
of  vessels,  as  the  half  bushel,  peck  measure,  pint  cup,  quart  cup,  gallon  mea- 
Bttre,  &c. 

The  kinds  of  measure  we  are  now  about  to  mention  are  used  to  measure  ex- 
tension;  thai  is,  distances,  surfaces,  Sfc.     For  this  purpose  we  haye  the  Sur- 


I^orwhatis  Beer  measure  used?     What  are  the  denominations?     Repeat  the 
TableT  '   How  many  cubic  inches  does  a  gallon,  Beer  measure,  contain?     For 


what  are  the  preceding  measures  used'? 
For  what  are  the  other  measures  used! 


What  do  we  have  for  this  purpose? 
What  do  we  have  for  this  purpose? 


158  COMMON   ARITHMETIC.  ScCt.  IX 

veyor's  chain,  the  square,  or  rule,  yard  measure,  tapes,  and  lines  of  differen 
lengths,  &c. 
2. — Extension  has  three  dimensions  :  length,  breadth,  and  thickness, 

1.    LONG  MEASURE. 

Obs.  10.  This  measure  is  used  when  length  or  distance  is  con- 
sidered, without  regard  to  breadth  or  thickness.  It  is  frequently  called 
Linear,  or  Lineal  measure. 

The  denominations  are  Inch,  Foot,  Yard,  Rod,  Furlong,  and 
Mile. 

TABLE. 

12    inches  (in.) make  1  foot, marked  ft. 

3    feet **        1  yard, ♦*     yd. 

5i  yards,  or  161  feet,..    <■  \    ^  ^;tle,r"''...°:        "     rd. 

40    rods,  or  220  yards,  _  >     *'       1  furlong, **      fur. 

8    furlongs,  or  320  rods,    *'       1  mile, .         **     M. 

ALSO  : 

3  barley  corns,  (bar.  c.)    '*       1  inch,  (but  little  used.) 

4  inches "\    I  I^and,  used  in  measuring  the 

I  height  of  horses. 

6    points "       1  line ;  and 

^2    1-  ,.  ^    1  inch,   used   in   measurinir  the 

6 


length  of  clock  pendulums. 
n    .  <i  S    ^  fathom,  used  in  measuring  the 

depth  of  water. 
2       .,  «'  ^    ^  league,  used  in  measuring  dis- 

^  tances  at  sea. 

''  ^StSsr.--!!S "    '  -^^g-' "  <^^r--°- 

ogQ    o  n  ^    t^G    circumference,    or   distance 

^  round  the  earth. 

Kemare. — The  standard  unit  of  length  adopted  by  the  United  States  is  the 
Yard  of  3  feet,  or  36  inches,  and  is  the  same  as  the  Imperial  Yard  of  Great 
Britain.  It  is  made  of  brass,  and  is  determined  at  the  temperature  of  60'^ 
Fahrenheit,  from  the  scale  of  Troughton,  a  celebrated  English  artist. 


For  what  is  Long  measure  used?  What  is  it  frequently  called?  What  are 
the  denominations?  Repeat  the  Table.  What  is  the  standard  unit  of  length 
adopted  by  the  United  States?     How  determined? 


*  This  i3  not  exactly  corre-t,  although  it  is  what  is  usually  given.  On  the  equator,  69  and 
one-sixth  statute  miles  make  1°,  or  60  geograpliical  milea,  iieirly ;  and  on  the  meridian,  at  a 
mean,  69^  statute  miles  make  1<\ 


Art.  2. 


REDUCTION. 


159 


1.  How  many  inches  in  2  feet?     3;  5;  7;  12;  8;  6;  9;   11;   15; 

2.  How  many  feet  in  3  yards?     4;  9;   12;  6;  6;  8;  22? 

3.  How  many  furlongs  in  2  miles?     4;  6;  9;   11;  8;   12? 

4.  How  many  inches  in  1  yard?     2;  6;  3;  4? 

6.  How  many  feet  in  24  inches?     48;   108;  36;  84;  60? 

6.  How  many  yards  in  6  feet?     12;  36;  24;   15;  33;   18? 

7.  How  many  miles  in  16  furlongs?     32f  72;  96;  48;  64? 

8.  How  many  yards  in  36  iuches?     108;   180;  72;   144? 

9.  How  many  barley-corns  in  a  foot?     2;  4;  6;  3? 

10.  How  many  feet  in  3  hands?     4;  7;  8;  5;  9;  6;  12? 

11.  How  many  points  in  1  inch?     2;  3;  4? 

12.  How  many  feet  in  2  fathoms?     3;  5;  9;   12;  8;  6;  7? 

13.  How  many  feet  in  36  barley-corns?     72;   144;  80;   108? 

14.  How  many  miles  in  2  leagues?  4;   12;  8;  5;  7;  8? 

15.  How  many  hands  in  12  inches?     1  ft.  4  in,;  3  ft;  2  ft,   8  in, 

16.  How  many  inches  in  72  points?     144;  216;  288? 

17.  How  many  fathoms  in  \2  feet?     24;  72;  28;  60,   24? 

18.  How  many  leagues  in  6  miles?     12;  36;  24.  32;  48? 

19.  How  many  iiwrfies  in  2  ft.  3  in.?  3  ft.  6  in.;  3  ft.  9  in.;  6  ft.? 

20.  How  many  feet  in  25  inches?     33;  40;  44;  57;  70? 


2 1 ,  Reduce  85  miles  to  inches. 

23.  Reduce  7  M.  6  fur.  27  rds, 
3  yds.  2  ft.  to  inches. 

25.  Reduce  7°.  49  M.  7  fur. 
16  rds.  12  ft.  9  in.  2  bar.  c.  to 
barley  corns. 

Note. — Statute  miles  are  uncer- 
Btood  in  this  example. 

27,  How  many  barley  corns 
would  it  take  to  reach  around  the 
globe,  it  being  360  degrees. 

Note. — Multiply  by  69i  to  reduce 
it  to  miles. 

29.  How  man,y  t'mes  will  a  car- 
riage wheel  turn  over  in  going 
36'J  miles,  it  being  16  ft.  6  in.  in 
circumference,  that  is,  around 
the  outside? 


22.  Reduce  5385600  inches  to 
miles. 

24.  Reduce  496518  inches  to 
miles. 

26.  Reduce  101964125  barley 
corns  to  degrees. 


28.  How  many  degrees  is  it 
around  the  globe,  it  being  4755- 
801600  barley  corns? 


30.  How  many  miles  does  a 
carriage  wheel,  16  ft.  6  in  in.  cir- 
cumference, proceed,  in  turning 
over  115200  times. 


2.     CLOTH  MEASURE. 

Obs^ll.     This  is  used  in  measuring  cloth,  tape,  lace,   and  all 
kinds  of  goods  bought  and  sold  by  the  yard. 


160 


COMMON    ARITHxMETIC. 


Sect.  IX 


The  denominations  are  Nail,  Quarter,  Yard,  and  Ell. 


TABLE, 


4  nails  (na.)  _ make  1  quarter,  __. 

4  quarters ''  1  yard, 

3  quarters,  or  J  of  a  yd.    "  1  Flemish  ell, 

5  quarters,  or   \\  yards     "  1  English  ell, 

6  quarters,   or   1^  yards    **  1  French  ell, 


marked  qr. 
"  yd. 
**      Fl.  e. 

*      E.e. 

'       F.  e. 


16;   11;  9? 

I 

5;   12;   111 

■-£ 

i 

Remark. — Cloth  measure  is  a  species  of  Long  measure.  Tlie  yard  is  the  same 
in  both  ;  that  is,  it  contains  3  feet,  or  36  inches.  Therefore,  a  quarter  contains 
9  inches,  and  a  nail  2^  inches.  Cloths,  &c.,  are  bought  and  sold  according  to 
their  length,  without  regard  to  their  width.  The  nail  is  but  little  used,  eighths 
and  sixteenths  being  used  in  its  place. 

1.  How  many  nails  in  2  quarters?     4;  8;   12; 

2.  How  many  quarters  in  2  yards?     9;  9;  7; 

3.  How  many  nails  in  2  yards?     3;  6;  4;  6? 

4.  How  many  quarters  in  2  Flemish  ells?     4;  8;  9;  6;  12?     ' 

5.  How  many  quarters  in  2  English  ells?     ^4;  7;  5;  9;   12?- 

6.  How  many  quarters  in  2  French  ells?     12;  9;  8;  7;  5;  6? 

7.  How  many  quarters  in  8  nails?      16;  48;  32;   12;  44;  36? 

8.  How  many  yards  in  8  quarters?     48;   12;   16;  32;  36;  44? 

9.  How  many  yards  in  6  nails?     64;  32;  80;  48;  96? 

iQ.  How  many  Flemish  ells  in  6  quarters?  36;   18;  30;  27;  52? 

11.  How  many  English  ells  in  10  quarters?  44;   16;  50;  20?  : 

12.  How  many  French  ells  in  12  quarters?  24;  4^;  54;  72;  48? 
13.  Reduce    6  yds.   3  qrs.  to 

nails. 

15.  Reduce  15  yds.  1  qr.  3  na. 
to  inches. 

17.  Reduce  8  F.  e.  3  qrs.  to 
nails. 

19.  Reduce  10  E.  e,  1  qr.  2 
na.  to  nails. 

21.  Reduce  12  F.  e,  toFl.  e. 

First  reduce  the  French  ells  to  quar- 
ters, and  then  to  Flemish  ells. 

23.  Reduce  25  F.  e.  to  En- 
glish ells. 

25.  Reduce  18  Fl.e.  to  French 
ells. 


14.  Reduce  108  nails  to  yards. 


555|    inches 


to 


16.  Reduce 
yards. 

18.  Reduce  204  nails  to  French 
ells. 

20.  Reduce  206  nails  to  En- 
glish ells. 

22.  Reduce  24  Fl.  e.  to  F.  e. 

Note. — First  reduce  the  Flemish 
ells  to  quarters,  and  then  to  French 
ells. 

24.  Reduce  30  E.  e,  to  French 
ells. 

26.  Reduce  9  F.  e.  to  Flem- 
ish ells. 


For  what  is  Cloth  measure  used?  What  are  the  denominations?  Repeat 
he  Table.  What  kind  of  meaiare  is  Cloth  measure?  What  measttxe  is  the 
ame  in  both?  How  many  feet  and  inches  does  the  yard  contain?  How  are 
loths,  &c.,  bought  and  sold? 


Art.  2. 


REDUCTION. 


161 


27.  Reduce  15  Fl.  e.  to 
Frencli  ells. 

29.  Reduce  36  E.  e.  to  Flem- 
ish ells. 


28,  Reduce  9  E.  e.  to  French 
ells. 

30.  Reduce  60  Fl.  e.  to  English 
ells. 


M 


3.    LAND,  OR  SQUARE  MEASURE. 


Obs.  12.  This  is  used  for  measuring  land ,  flooring ,  or  anything 
in  which  length  and  breadth  are  considered,  withovi  regard  to  depth  or 
thickness. 

The  denominations  are  Square  Inch,  Square  Foot,  Square  Yard, 
Square  Rod,  Rood,  Acre  and  Square  Mile, 


A.  Obs.  13.  When  any  two  lines  meet  together,  the  open- 
ing at  their  pilace  of  meeting  is  called  an  Angle. 
When  they  meet  so  as  to  form  a  square  corner, 


the  corner  A.,  the  angle  is  called  a  Right 


like 
Angle. 


1  square  yard. 


3 

ft.  lou 

3 

Obs.  14.  A  figure  having  four  equal 
sides,  and  its  angles  right  angles,  is  called  a 
Square. 

A  Square  inch  is  a  square,  each  side  of 
which  measures  an  inch  in  length. 


A  Square  yard  is  a  square,  each  side  of  which  is  a  yard,  or  3  feet 
in  length,  and  contains  9  square  feet.  A  square  foot  contains  12  X 
12=  1 44  square  inches. 


TABLE. 


144    square  inches{sq.  in.)  make  1  square  foot,  _ _ 

9    square  feet "       1  square  yard, 

30^^  square  yards, or  272^  >  <(    S^  square  rod,  perch, 

square  feet, )        (  or  pole, 

40    square  rods "       1  rood, ; 

4    roods,  or  160  sq.  yds.     **       1  acre, •_ 

^^     i  1  square    mile,    or 
^  section, 


640    acres 


marked  sq.  ft. 
sq.yd^ 

sq.rd' 

*      R. 
A. 

sq.M. 


For  what  is  Land,  or  Square  measure  used?  What  are  the  denominations? 
What  is  sin  angle?  A  right  angle?  A  square?  A  square  inch?  A  square 
yard?  llf.w  many  square  feet  does  a  square  yard  contain?  How  many 
square  inches  does  a  square  foot  contain?    Repeat  the  Table. 


162 


COMMON    ARITHMETIC. 


Sect.  IX. 


In  measuring  land,  surveyors  use  a  chain  4  rods  long,  and  con* 
tainining  100  links.     Hence — 

7_.y_.  inches make  1  link, marked  1. 

25        links "       1  rod, "      rd. 

4        rods  or  66  feet, **       1  chain, *'      ch. 

80        chains- "       1  mile, **      M. 

These,  the  learner  will  observe,  are  all  linear  measure.     But 

1  square  chain makes  1 6  perches, marked  P, 

10  square  chains make      1  acre, **      A. 

This  chain  is  generally  called  Gunter's  Chain,  from  the  name 
of  its  inventor. 

1.  How  many  square  feet  in  2  square  yards?     3;  6;  12;  9;  7? 

2.  How  many  roods  in  2  acres?     4;  9;  5;  7;  9;  12;  10? 

3.  How  many  square  yards  in  18  square  feet?     27;  54;  90? 
3.  How  many  acres  in  8  roods?     44;  32;  24:  48;  36;  49? 


^  Reduce  1  sq.  M.  326  A.  3 
R.  27  sq.  rd.  16  sq.  yds.  to  square 
inches. 

7.  Reduce  16  A.  2  R.  17  sq. 
rd.  124  sq.  ft.  127  sq.  in.  to 
square  inches. 

9.  Reduce  7  A.  120  sq.  rd. 
212  sq.  ft.  114  sq.  in.  to  square 
inches. 

11.  If  a  man's  farm  measures 
120  chains  in  length,  how  many 
rods  long  is  it?  How  many 
miles? 


6.  Reduce  6065153964  sq.  in. 
to  square  miles.  ? 


8.  Reduce 
to  acres. 


104183011  sq.  in. 


10.  Reduce  48643602  sq.  in. 
to  acres. 

12.  If  a  man's  farm  measures 
\\  miles  in  length,  how  many 
rods  long  is  it?  How  many 
chains? 


The   student  will  observe  that  the 
mile  isLoog  measure. 

Obs.  15.     Although  a  square  foot  and  2k  foot  square  are  the  same, 
there  is  a  diflference  between  square  feet  and/ee/  square. 

3  ft.  sq.=9  sq,  ft. 


It  will  be  perceived  from  the  figure,  that  3 
feet  square  measures  3  feet  on  -each  side,  and 
contains  3x3  =  9  square  feet. 


What  is  used  in  measuring  land?  How  long  is  this  chain?  How  many 
links  does  it  contain?  Repeat  the  Table.  What  name  is  usually  given  to  thi« 
chain?     Why? 


'Art.  2. 


REDUCTION. 


163 


3  pq.  ft 

1 

On  the  other  hand,  3  square  feet  measures  3 
feet  in  length,  and  only  1  foot  in  width,  and 
contains  3X1=3  square  feet.  Therefore, 
there  is  a  difference  of  6  sq.  ft.  between  3  ft.  sq.  and  3  sq.  ft. 

Between  2  sq.  ft.  and  2  ft.  sq.  there  is  a  difference  of  2  sq.  ft. 
Between  4  sq.  ft.  and  4  ft.  sq.  there  is  a  difference  of  12  sq,  ft. 

13.  What  is  the  difference  between  8  sq.  ft.  and  8  ft.  sq.? 

14.  What  ts  the  difference  between  9  sq.  ft.  and  9  ft.  sq,? 

15.  What  is  the  difference  between  12  sq.  ft.  and  12  ft.  sq.? 

16.  What  is  the  difference  between  25  sq.  ft.  and  25  ft.  sq. 

Ans.  600  sq.  ft. 

Besides  the  square,  we  have  other  four-sided  figures,  which  have 
different  names,  according  to  their  forms.  The  most  common  are 
the  Rectangle,  Parallelogram,  and  Rhombus. 


Rectangle. 


Parallelogram. 


QO 

/// 

/    //- 

«f    // 

'  /  M 

4  ft.  Ions. 


4  ft.  long. 


Obs.  16.  The  space  enclosed  by  the  lines  which  bound  a  figure, 
is  called  its  Area,  or  superficial  contents. 

It  will  be  perceived  that  the  area  of  the  above  figures,  as  well  as 
as  that  of  the  square,  is  found  by  multiplying  together  their  length 
and  breadth.    Hence — 

To  find  the  area  of  a  square,  rectangle,  parallelogram,  or  rhom- 
bus : 

Obs.  1 7.     Multiply  together  their  length  and  breadth  : 

Remark. — Recollect  that  feet  multiplied  by  feet  produce  square  feet,  and  so 
of  all  the  other  denominations  of  linear  measure. 

17.  What  is  the  difference  in  the  size  of  two  rooms,  one  being  15 
feet  square,  and  the  other  containing  1 96  square  feet? 

Ans.  The  one  1 5  feet  square  contains  29  sq.  ft.  the  most. 


What  is  the  difference  between  a  square  foot  and  a  foot  square?  Be- 
tween SV^.ft.  and  3  ft  sq.?  Show  why  this  is  the  case.  What  four  other  fig- 
ures have  we  besides  the  square?  What  is  the  area  of  a  figure?  How  do  we 
find  the  area  of  a  square,  rectangle,  parallelogram,  or  rhombus?  What  do 
the  denominations  of  linear  measure,  multiplied  by  the  same,  produce? 


161  COMMON    ARITHMBTIC.  ScCt.    IX. 

18.  If  a  floor  is  18  feet  long,  and  12  feet  wide,  how  many  square 
feet  does  it  contain?  Ans.  216. 

19.  If  a  piece  of  land  is  45  chains  long,  and  80  rods  wide,  how 
many  acres  does  it  contain?  Ans.  90. 

Reduce  both  to  the  same  denomination. 

20.  How  many  square  feet  in  a  board  18  inches  wide,  and  24  feet 
long?  Ans.  36. 

Reduce  both  to  the  same  denomination. 

21.  How  many  square  feet  in  a  board  32  feet  long,  and  16  inches 
long?  Ans.  42  sq.  ft,  96  sq.  in.     . 

22.  How  many  acres  in  a  square  field,  each  side  of  which  mea- 
sures 120  rods?  ■  Ans.  90. 

23.  How  many  acres  in  a  field  46  rods  long,   and  20  rods  wide? 

Ans,   5  A.  120  sq.  rds. 

24.  The  largest  of  the  Egyptian  pyramids  is  a  square  at  the  base, 
each  side  being  about  693  feet  in  length.  How  many  acres  does  it 
cover?  Ans.  11  A.  4  sq.  rds. 

25.  What  is  the  difference  between  a  floor  25  feet  square,  and 
two  others,  each  16  feet  square?  Ans.   113  sq.  ft, 

26.  If  a  room  is  30  feet  long,  how  wide  must  it  be  to  contain  240 
sq.  ft.?  Ans.  8  feet. 

Note. — The  length  and  breadth  are  the  factors,  which,  multiplied  together, 
produce  the  area.  Therefore,  the  area  divided  by  either  factor,  will  give  the 
other.     (Sect.  VI.  Art.  1.  Obs.  16.) 

27.  If  a  board  is  18  inches  wide,  hoAV  long  must  it  be  to  contain 
18  sq.ft.?  Ans.  12  ft. 

Reduce  both  to  the  same  denomination. 

28.  If  a  board  is  15  feet  long,  how  wide  must  it  be  to  contain  20 
sq.  ft.?  Ans.   16  in. 

29.  If  a  room  is  12  ft.  long,  how  wide  must  it  be  to  contain  216 
sq.  ft-?  Ans,   18  ft. 

30.  How  many  yards  of  muslin,  J  of  a  yard  wide,  will  it  take  to 
line  3  yds.  satinet,  Ij  yds.  wide?  Ans.  5  yds. 

4.    SOLID,  OR  CUBIC  MEASURE. 

This  is  used  to  measure  wood,  stone,  or  any  thing  in  which 
length,  hreadth,  and  thickness,  or  depth,  are  considered. 

The  denominations  are  Solid  Inch,  Solid  Foot,  Solid  Yard, 
Cord,  and  Ton. 

Obs,  19.  A  solid  body  having  six  equal  square  faces,  is  called  a 
Cube,  or  Hex.^dron. 

When  we  have  the  area,  and  either  the  length  or  breadth  given,  how  do  we 
find  the  other?  Why  is  this  correct?  For  what  is  Solid  or  Cubic  measure 
used?    What  are  the  denominations? 


Art.  2. 


REDUCTION. 


161 


3ff. 


If  each  side  of  a  cube  measures 
an  inch  in  length,  it  is  called  a  cubic, 
or  solid  inch.  If  each  side  measures 
a  yard  in  length,  it  is  called  a  cubic, 
or  solid  \jard,  and  contains  27  solid 
feet ;  that  is,  3  feet  in  length  3  feet 
in  width,  and  3  feet  in  thickness. 

A  solid  foot  contains  12  X  12  X 
12=  1728  solid  inches. 


Note. — This  can  best  be  explained  by  a  number  of  small  cubical  blocks, 
with  which  any  teacher  oan  supply  himself. 


TABLE. 

make 


1  solid  foot, marked  s.  ft. 

1  solid  yard, **      s.yd. 

1  ton, "      T. 

1  cord  of  wood  or 
bark,  . ''      C. 


1728  solid  inches  (s.  in.)_. 

27  solid  feet 

50  feet  of  round,  or 
40  feet  of  hewn  timber. 
128  solid  ft.  =:  8  X  4  X 
4,  that  is  8  ft.  long,  4 
ft.  wide,  and4  ft.high,  ^ 

Remark  1. — A  pile  of  wood  1  foot  long,  4  feet  wide,  and  4  feet  high,  is 
called  a  cord  foot,  and  contains  16  solid  feet.     8  cord  feet  make  1  cord. 

2. — In  estimating  the  tonnage  of  ships,  42  solid  feet  are  allowed  for  a  ton. 

3. — By  a  ton  of  round  timber,  is  meant  such  a  quantity  of  timber  in  its 
rough,  or  natural  state,  as  when  hewn  will  make  40  cubic  feet. 

4. — A  cubic  foot  contains  7.43  wine  gallons,  and  6.127  beer  gallons. 

5. — A  cubic  foot  of  distilled  water  weighs  about  1000  oz.  Avoirdupois,  or  very 
nearly  624  lbs.,  at  40^^  temperature.  At  60'^,  which  is  generally  used,  it  weighs 
only  "62.353  lbs.,  which  is  less  than  1000  oz.  The  foot  weighs  911.458  oz. 
Troy,  or  .5274  oz.  per  cubic  inch. 


1.  Reduce  6  solid  yards  to 
solid  inches. 

3.  Reduce  12  tons  of  round 
timber  to  solid  inches. 


2.  Reduce  279936  solid  inches 
to  solid  yards. 

4.  In  1036800  solid  inches, 
how  many  tons  of  round  timber? 


What  is  a  cube?  A  cubic  inch?  A  cubic  yard?  How  many  solid  feet  does 
a  solid  yard  contain?  How  many  solid  inches  in  a  solid  foot?  Repeat  the 
Table.  What  is  a  cord  foot?  How  many  solid  feet  does  it  contain?  How 
many  cord  feet  make  a  cord?  How  many  solid  feet  are  allowecl  for  a  ton  in 
estimating  the  tonnage  of  vessels?  What  is  meant  by  a  ton  of  round  timber? 
How  many  wine  gallons  does  a  cubic  foot  contain?  Beer  gallons?  What  is 
the  weight  of  a  cubic  foot  distilled  water?  How  do  we  find  the  contents  of  a 
solid,  the  sides  of  which  are  squajes,  rectangles,  &c.? 


165  COMMON     ARITHMETIC.  '       Sect.  IX. 


5.  In  5  cords  of  wood,  how 
many  cord  feet?  How  many 
solid  feet? 

7.  A  ship  contains  12600  solid 
feet.     Required — the  tonnage. 


6.  In  540  solid  feet  of  wood, 
how  many  cord  feet?  How  many 
cords? 

8.  Required — the  number  of 
solid  feet  in  a  ship  of  300  tons,  h 


From  the  remarks  under  Obs.  17,  we  conclude  that  f  . 

To  find  the  contents  of  a  solid,  the  sides  of  which  are  squares,'" 
rectangles,  <fec. :  a 

Obs.  20.  MuU'qyly  together  its  length,  breadth,  and  thickness,  or 
depth. 

9.  How  many  solid  feet  in  a  room  18  feet  long,  16  feet  wide,  and 
12  feet  high?  Ans.  3456. 

10.  How  many  solid  feet  in  a  box  3  feet  long,  2  feet  wide,  and 
60  inches  deep?  Ans.  8. 

Reduce  all  to  the  same  denomination. 

11.  How  many  cubic  feet  in  a  pile  of  wood  14  feet  long,  4  feet 
wide,  and  3~  feet  high?  Ans.  196. 

12.  How  many  cords  of  wood  in  a  pile  186  feet  long,  6  feet 
wide,  and  A\  feet  high?  Ans.  39  cords,  1  j  cord  feet. 

5.    TIME. 

Obs.  21.  T iifne  is  the  measure  of  duration.  Ifc  is  naturally  divided 
into  days  and  years,  the  former  being  caused  by  the  revolution  ot 
the  earth  around  its  axis,  and  the  latter  by  revolution  around  the 
sun. 

Other  divisions  have  been  made  by  man  ;  the  day  being  divided 
into  hours,  minutes,  and  seconds ;  and  the  year  into  months,  weeks, 
and  days. 

TABLE. 

60  seconds  (sec.) make    1  minute ....  marked  min. 

60  minutes "      1  hour. ♦*  h. 

24  hours-- "      1  day "  d. 

36  5|^  days "      1  year **  yr. 

100  years *'      1  century "        cen. 

ALSO  : 

7  days makes  1  week marked  wk. 

4  weeks **       1  month **       mo. 

13  months,  1  day,  and  6  hours,  (nearly,)  or  365;|  days,  makes  1 
common  or  Julian  year. 

How  is  Time  naturally  divided?  What  causes  the  day?  The  year?  How 
fc  the  day  divided?    The  year?    Repeat  the  Table. 


Art.  2.  REDUCTION.  167 

Remark  1. — A  Solar  year  is  the  exact  time  in  which  the  earth  revolvet 
around  the  sun,  and  contains  365  d.  5  h.  48  min.  48  sec. 

2.  The  6  hours,  or  :^  of  a  day  is  not  added  to  each  year,  but  reserved  until 
every  fourth  year,  when  a  whole  day  is  gained.  Therefore,  every  fourth  year 
must  contain  366  days.     This  year  is  called  Bissextile,  or  Leap  Year. 

3.  Leap  Years  are  those  which  can  be  divided  by  4  without  a  remainder;  as 
1840,  1844,  1848,  &c. 

4.  Exceptions. — It  will  be  perceived  that  in  counting  365^  days  to  the  year, 
we  reckon  11  min.  and  12  see.  too  much.  This  in  100  years  amounts  to  I8hrs. 
40  min.;  dropping  the  40  min.,  it  being  too  small  to  be  noticed,  we  fiud  that 
each  lOO  years  we  count  18  hrs.  or  |  of  a  day  too  much,  and  in  400  years  we 
increase  3  days  in  time.  Therefore,  to  produce  the  exact  time,  we  count  every 
fourth  centurial  year  as  leap  year,  and  the  intermediate  centurial  years  as  com- 
mon years,  (365  days.)  Hence — If  the  centurial  years  can  be  divided  by  400 
they  are  leap  years,  otherwise  they  are  not. 

Examples.— 1200,  1600,2000,  &c.,are  leap  years,  but  1500, 1700,  1800,  &c. 
are  not. 

5.  The  year  is  also  divided  into  12  calendar  months,  the  order  of  which,  and 
the  number  of  days  in  each,  are  as  follows: 

January,     (Jan.)    1st  month  has  31  days. 

February,  (Feb.)    2d  «'  *«  28  " 

March,        (Mar.)    3d  "  «  31  «* 

April,          (Apr.)   4th  *«  «  30  " 

May,                          5th  "  «  31  « 

June,                        6th  "  «  30  " 

July,                         7th  "  «  31  " 

August,      (Aug.)  8th  "  "  31  '* 

September,  (Sept.)  9th  «  «  30  « 

October,       (Oct.)   10th  «  «  31  " 

November,  (Nov.)  11th  "  "  30  " 

December,  (Dec.)  12th  "  "  31  " 

6.  The  odd  day  which  is  added  to  every  fourth  year,  is  added  to  the  month 
of  February,     Therefore,  every  fourth  year  February  has  29  days. 

7.  The  number  of  days  in  each  month  may  be  easily  remembered  by  com- 
mitting to  memory  the  following  lines: 

Thirty  days  hath  September, 

April,  June,  and  November  ; 

All  the  rest  have  thirty-one. 

Save  February  alone  ; 

Which  hath  twenty-eight  in  store, 

Till  Leap  year  gives  it  one  day  more. 

1.  How  many  days  in  2  weeks?     5?  6?  4?  7?  3?  9?  8?  125 

2.  How  many  weeks  in  2  months?     4?  6?  3?  7?  5?  8?  9?  12? 

3.  How  many  minutes  in  2  hours?     3?  6?  4?  7?  5?  8?  12?  9? 

What  is  a  solar  year?  What  is  its  length  ?  What  is  done  with  the  6  hours 
or^  of  a  day?  How  many  days  has  every  fourth  year?  What  is  this  year 
called?  Which  are  leap  years?  Give  examples.  How  is  the  year  otherwise 
divided?  Repeat  the  names  of  the  months  together,  with  the  number  of  days 
in  each.  What  is  done  respecting  the  centurial  years?  To  which  month  is 
(he  odd  day  added?  How  many  days  j^then  has  February  in  leap  year?  Re* 
peat  the  lines  for  remembering  the  number  of  4^yg  in  e^fh  mouth? 


1G8 


COMMON    ARITHMETIC. 


Sect.  IX 


4.  How  many  weeks  in  14  days?     29?  42?  35?  56^  84? 

5.  How  many  months  in  8  weeks?     24?  48?  16?  40?  32?  44? 

6.  How  many  hours  in  120  minutes?     300?  720?  360?  480? 


7.  Suppose  a  person's  age  to  be 
18  yrs.,  217  df,  16  h.,  43  min.. 
37  sec  ,  how  many  seconds  old  is 
he? 

9.  How  many  seconds  in  3 
centuries,  57  yrs.,  7  mo.,  3  wks. 
1  d.,  18  h.,  33  min.? 

11.  How  many  seconds  be- 
tween July  4th.,  1776,  and  July 
4th.,  1848,  it  being  72  years? 


8.  Reduce  586845817  seconds 
to  years. 


10.  Reduce  10382754780  sec- 
onds to  centuries. 

12.  Reduce  2272147200   sec- 
onds to  years. 


6.    CIRCULAR  MEASURE,  OR  MOTION. 

Obs.  22.     This  is  used  in  estimating  latitude  and  longitude, 
also  in  measuring  the  motions  of  the  heavenly  bodies. 


and 


All  the  calculations  by  this  measure  are  made  in  circles,  every 
circle,  whether  large  or  small,  being  supposed  to  be  divided  into 
360  equal  parts  called  degrees.  Each  degree  is  divided  into  60 
equal  parts  called  minutes,  and  each  minute  into  60  equal  parts 
called  seconds. 


TABLE. 


60  seconds  (") make  1  minute marked 

60  minutes "       1  degree '* 

30  deo-rees 


12Signs,  or  360°. 


1  sign  _ 
1  circle 


c. 


Remark  1. — The  Sign  is  but  little  used,  calculations  by  this  measure  being 
chiefly  made  in  degrees,  minutes,  andseconds^ 

2.  The  minute  is  the  same  as  the  geographical  mile,  60  of  which  make  a  de- 
gree. 

3.  It  must  not  be  inferred ,  however,  that  60  geographical,  or  69i  satuto  miles 
(table  long  measure,)  make  a  degree  in  every  circle.  This  is  only  the  case 
when  the  circumference  of  the  circle  is  the  same  as  the  circumference  of  the  earth. 
If  the  circle  is  larger  or  smaller,  the  distance  is  greater  or  less  in  proportion. 


For  what  is  circular  measure  used?  How  are  the  calculations  made?  How 
is  the  circle  divided?  The  degree?  The  minute?  Repeat  the  Table.  What 
denominations  are  used  principally  in  calculating  by  this  measure?  To  what 
does  the  minute  correspond?  How  large  must  a  circle  be  to  have  6)  geograph- 
ical, or  69i  statute  miles  make  a  degree?  ***  *'*^"  "* 


Art.  2. 


REDUCTION. 


169 


1.  How  many  signs  in  2  circles?     4?  6?  10?  8?  5?  7?  12? 

2.  How  many  degrees  in  2  signs?     4?  5?  7?  9?  8?  12? 

3.  How  many  minutes  in  2  degrees?     4?  8?  5?  7?  6? 

4.  How  many  circles  in  24  signs  ?     48?  60?  36?  72?  144? 

5.  How  many  signs  in  60  degrees?     120?  210?  180?  240? 

6.  How  many  degrees  in  120  minutes?     180?  300?  540?  720? 


7.  Reduce  17°  43'  ST  to  sec- 
onds, 

9.  Reduce  8s.  29°  39"  to  sec- 
onds. 

II.  Reduce  Ic.  355°  59'  49" 
to  seconds. 


8.  Reduce  63837"  to  degrees 

10.  Reduce  968439"  to  signs. 

12.  Reduce   2577589"  to  cir- 
cles. 


MISCELLANEOUS    TABLE. 


1 2  things make  1  dozen marked  doz. 


12  dozen. 
12  gross. 


20  tilings 

5Q  pounds. 

112  pounds 

1 96  pounds 

200  pounds  . .  _  ,     

14  pounds  of  iron  or  lead 

21-^  stone 

8pigs 

24  sheets  of  paper 

20  quires 


1  gross <* 

1  great  gross ** 

1  score. 

1  firkin  of  butter.  _       ** 

1  quintal  of  fish  _  _        " 

1  barrel  of  flour. 

1  barrel  ol  pork,  or  beef. 

1  stone. 

1 

1 

1  quire. 

1  ream. 


g- 


gi-- 
gr- 


fir. 
quin. 


pig- 
fother. 


BOOKS. 

A  sheet  folded  in  two  leaves  is  called  a  folio. 


A  sheet 
A  sheet 
A  sheet 
A  sheet 


four 
eight 
twelve 
eighteen 


a  quarto,  or  4  to. 
an  octavo,  or  8  vo, 
a  duodecimo,  or  12  mo. 
an  1 8  mo. 


how 


MISCELLANEOUS   EXERCISES    FOR    THE    SLATE, 

Involving  the  prmciples  of  Reduction. 
1.  In  15  bars  of  silver,  each  weighing  6  lbs.  8  oz.  15pwt8. 1 1  grs., 


many  grams! 


2.  How  many  thimbles,  each  weighing  8  pwts. 
made  from  1  lb.  8  oz.  16  pwts.  of  silver? 


Ans.  581565. 
16  grs.  can  be 
Ans.  48. 


Repeat  the  Miscellaneous  Table.     What  is  a  sheet  folded  in  two    leave* 
called?    In  four  leaves?    In  eight  leaves?    In  12  leaves?    InlSleavtM? 
9 


]70  COM^ION    ARITHMETIC  ScCt.   IX 

3.  How  many  rings,  each  weighing  5  pwts,  7  grs.,  can  be  made 
from  2  lbs.  8  oz.  16  pwts.  4  grs.  of  gold?  Ans.   124. 

4.  How  many  pounds  of  hay  in  6  loads,   each  weighing  14  cwt. 
2  grs.  18  lbs.?  Ans.  8808. 

5.  How  many  nails,  each  weighing  6  drams,  can  be  made  from 
29  lbs.  3oz.  4drs.  o'  iron?  Ans.   1246. 

6.  How  many  casks,  each  holding  68  lbs.,  can  be  filled  from  16 
cwt.  1  gr.  7  lbs  of  sugar?  Ans.  24. 

7.  If  a  family  consume  4  lbs.  12  oz  of  coffee  in  a  month,  how  long 
will  1  cwt.  3  qrs.  9  lbs.  8oz.  last  them?  Ans.   38^4.  months. 

8.  In  45  pounds  avoirdupois,  how  many  pounds  troy? 

Ans.  54\\. 

9.  In  36  pounds  trov,  how  many  pounds  avoirdupois? 

Ans.  29V^|. 

10.  A  physician  made  136  pills,  each  containing  4  grains,  how 
much  did  they  all  contain?  Ans,   I3,  I3.  4  grs. 

11.  How  many  doses  of  calomel  each  containing  8  grs,,  can  be 
made  from  lib,  23,  33.  29,  12  grs.  Ans,  869. 

12.  What  quantity  of  rhubarb  will  it  take  to  put  up  324  doses, 
each  containg  24  grs.  Ans.   lib.  43.  I3,  19,  16  grs. 

13.  A  merchant  wishes  to  put  675  bushels  of  clover  seed  into 
casks,  each  holding  6  bushels,  3  pecks.  How  many  casks  are  re- 
quired? Ans.   100, 

14.  At  i-  of  a  dollar  a  peck,  what  would  18  bu,  1  pk.  of  wheat 
cost?  Ans.   18^  dollars. 

16.  How  many  bottles  will  it  take,  each  holding  |^  a  pint,  (2  gills,) 
to  put  up  a  hogshead  of  wine?  Ans.  1008, 

16,  A  certain  cistern  holds  66  barrels  of  water,  how  many  times 
will  it  fill  a  pail  holding  3  galls,  1  qt.  1  pt,  Ans,  616  times, 

17,  How  many  casks,  each  holding  7  galls,  3  qts,  1  pt.,  can  be 
filled  from  a  tun  of  wine?  Ans.  32. 

18.  How  many  dozen  bottles,  each  dozen  containing  3  gal.  3  qts, 
1  pt.  2gi.,  can  be  filled  from  a  hogshead  of  cider?  Ans,   16. 

19,  How  many  gallon,  quart,  pint,  and  gill  bottles,  of  each  an 
equal  number,  can  be  filled  from  210  galls,  3  qts.  1  pt,  2  gi.  of  wine? 

Ans.   150. 

20.  How  many  pints  in  a  hogshead  of  ale?  Ans.  432. 

21.  How  many  gallon,  quart,  and  pint  bottles  will  it  take  to  put 
Tip  a  barrel  of  beer,  having  the  same  quantity  in  each  kind  of  bot- 
tles? 

Ans.  12  gallon  bottles,  48  quart  bottles,  and  96  pint  bottles. 

2£.  How  many  ccislis,  each  holding  6  gali<>.  3  qts.  1  pt.,  will  it 
take  tj  hold  19  hhds*,  19  galls  of  ale?  Ans.  1'52. 


Art.    2.  REDUCTION.  171 

23.  If  a  man  drink  1  pint  of  beer  each  day  for  24  years,  how 
much  would  it  amount  to?  Ans.  20  hhds.,  15  galls.,  3  qts. 

24.  How  many  quarters  in  12  pieces  of  cloth,  each  containing  15 
yds.,  2  qrs.         "  Ans.  744  =  186  yds. 

25.  How  many  pieces  of  cloth  each  containing  12  yds.  3  qrs.  2 
na.,  are  there  in  579  yds.,  1  qr.,  2  na?  Ans.  45. 

26.  How  many  suits  of  clothes,  each  containing  4  yds.  1  qr.,  3 
na.,  can  be  mad;i  from  53  yds.,  1  qr.  of  cloth?  Ans.   12. 

27.  How  many  pair  of  pantaloons,  each  conta'ning  1  yd.  2  qrs. 
3  na.,  can  be  made  from  18  Flemish  ells  of  broadcloath? 

Ans.   8. 

28.  How  many  times  will  a  carriage  wheel  16  ft.  6  in.  in  circum- 
ference, turn  over  in  going  1 8  miles?  Ans.  57b0. 

29  How  many  steps  of  2  ft.  5  in.  each,  ifill  it  a  man  take  in  trav- 
eling 1  m.  7  fur.  33  rods.  3  yds.  1  ft.?  Ans.  4326. 

30.  Ha  man  travel  20  m.  6  fur.  26  rds.  11  ft.  in  a  day,  how  long 
will  it  take  him  to  travel  around  the  earth,  it  being  about  25000 
miles?  Ar.s.   1200  days  =  3yrs.,  104  d.,  6  hrs. 

31.  How  many  times  will  a  ship.  133  ft.  4  in.  long,  sail  her  length 
in  crossing  the  Atlantic  ocean,  it  being  3000  miles. 

Ans.   118800. 

32.  How  many  yards  of  carpeting  I  yard  wide,  will  it  take  to 
cover  the  floor  of  a  room  16  feet  long,  and  12  feet  wide? 

Ans.  21|-. 

33.  A  land  owner  has  in  one  place  329  acres,  in  another  place 
870  acres,  in  another  place  1236  acres,  and  in  another  place  1405 
acres  ;  if  it  was  all  together,  how  many  square  miles  would  it 
make?  Ans.  6. 

34.  A  farmer  wishing  to  know  the  length  of  his  farm,  found  it  to 
be  146  chains,  66|  links.     Required  its  length  in  rods?     In  miles? 

Ans.  586|  rods  =  1|-  miles. 

35.  The  surface  of  the  planet  Venus  contains  about  9327776422- 
03545600  square  inches  ;  how  many  square  miles? 

Ans.  232352736. 

36.  How  many  blocks  3  inches  long,  and  2  inches  wide,  can  be 
cut  from  aboard  3  feet  In  length,  and  2  feet  in  width? 

Ans,   144.* 

37.  Apiece  of  land  is  144  rods  long,  and  95  rods  wide  ;  how 
many  acres  does  it  contain?  Ans.  85|-. 

38.  A.'s  farm  is  66  chains,  50  links  in  length,  and  45  chains,  75 
links  in  widih  ;  required  the  number  of  acres  owned  by  A.? 

Ans.  30411. 
■  ^  .3D.     If  a  roomis^lS  itj/et  log,  aud  12  feet  wide,  how  many  square 
Coet  dges  it  contain?****!***  >*»?  ^^  mJg^am.mLA-^.  Ans.  192. 


172  COMJUON    ARITHMETIC.  Scct.   IX 

40.  A  board  is  1 4  feet  long,  and  1 6  inches  wide  at  one  end,  and 
8  inches  wide  at  the  other  end  ;  required  the  number  of  square  feet 
in  the  board?  Ans.   14. 

The  width  at  one  end  is  16  inches,  and  at  the  other  end  8  inches  ; 
therefore  the  width  at  both  ends  is  16  -j-  8  =  24,  and  the  half  of 
this  or  12  inches,  nnui  be  the  mean  or  average  width.  12-r- 12  = 
1  foo^     14  X  1  ==  14  Ans.     Hence — 

To  find  the  mean  width  of  a  board,  or  anj  other  surface  whose 
edges  are  straight,  but  the  width  at  the  ends  are  different: 

Obs.  23.  Add  tor/ether  the  ivkUh  of  the  tioo  ends,  and  take  half  their 
sum. 

41.  If  a  board  is  24  feet  long,  and  32  inches  wide  at  one  end,  and 
18  inches  wide  at  the  other  end,  how  many  square  feet  in  the  board? 

Ans.  60. 

42.  How  many  square  feet  in  the  surface  of  a  rafter  19^  feet  long, 
3  inches  thick,  and  the  width  at  the  ends  4  and  3  inches? 

Ans.  21  sq.  ft.,  18  sq.  in. 

43.  There  is  a  room  18  feet  in  length,  16  feet  in  width,  and  8 
feet  in  heighth  ;  how  many  rolls  of  paper,  2  feet  wide,  and  contain- 
ing 1 1  yards  in  each  roll  will  it  take  to  cover  the  walls. 

44.  How  many  blocks,  2  inches  long,  2  inches  wide,  and  2  inches 
thick,  can  be  cut  from  a  cubical  blo::k  measuring  3  feet  on  each  side, 
allowing  no  waste  for  cutting?  Ans.  5832. 

45.  How  many  cords  of  wood  in  a  pile  56  feet  long,  12  feet  wide, 
and  8  feet  high?  Ans.  42. 

46.  A  cubic  post  of  oak  weiglis  950  ounces  avoirdupois ;  how 
many  pounds  will  6  cords  of  oak  weigh?  Ans.  45600. 

47.  If  a  clock  tick  60  times  a  minute,  how  many  times  will  it 
tick  in  a  year?  Ans.  31557600. 

48.  How  many  seconds  longer  was  March  than  February  in  the 
year  1847,  Ans.  259200. 

49.  What  are  the  leap  years  between  1846  and  1862? 

Ans.   1848,  1852,  1856,  1860, 

50.  If  a  person  wastes  J  of  an  hour  each  day,  how  much  time 
will  he  lose  in  28  years.  Ans.  319  d.  14  h.  15  min. 

51.  How  much  time  will  a  person  gain  in  50  years  by  rising  \  an 
hour  (30  min.)  earlier  than  usual,  each  day? 

Ans.   1  yr.  15  d.  5h.  15  min. 

52.  Sound  moves  at  the  rate  of  1142  feet  in  a  second  of  time; 

How  do  we  find  the  number  ©f  square  feet  in  a  board,  or  any  other  surface* 
wbos'e  edges  are  strait,  but  the  width  of  the  ends  different?  vv- 


Art.    2.  REDUCTION.  173 

required  the  time  it  would  take  sound   to  pass  from  the  sun  to  tlie 
planet  Uranus,  it  being  distant  IGOOuOOOOO  miles? 

Ans.  26  J  yrs.  201  d.  A  hr.  48  min.4- 

NoTE. — The  learner  will  observe  that  the  remainder  after  dividiug  by  3654  'S 
fourths  of  days,  not  whole  days.   ,, 

53.  If  a  person  spends  8  hours  of  the  day  in  sleeping  and  eating, 
how  many  minutes  per  day  does  he  devote  to  other  business. 

Ans.  960. 

54.  If  a  vessel  sails  10  miles  an  hour,  how  far  will  she  sail  in  4 
wks.  5d.  16hr3.  Ans.  8033  miles, 

55.  The  precise  length  of  the  tropical  year  is  365  d,  5  h,  48  min. 
48  sec;  how  many  such  years  in  1104492480  seconds? 

Ans.  35. 

56.  How  many  seconds  from  the  commencement  of  the  Christian 
Era,  until  Oct  7th,  1849,  4  o'clock,  P.  M.,  allowance  being  mode  for 
leap  year?  Ans.  58341398400. 

57.  If  a  planet  move  through  an  area  of  3°  36'  in  one  day,  how 
long  will  it  take  it  to  move  through  an  area  of  9s.  160°? 

Ans.  79|  days. 

58.  How  long  would  it  take  a  planet  to  move  through  a  quadrant 
(90°)  at  the  rate  of  59'  8"  per  minute?     Ans.   1  hr.,  31§||  min, 

59.  How  long  would  it  take  a  comet  to  move  through  a  semi- 
circle, (180°)  at  the  rate  of  8°  16'  daily?  Ans.  21|i  days. 

60.  What  cost  3  dozen  dozen  yards  of  cloth,  ai  2  dollars  a  yard? 

Ans.   864  dollars. 

6 1 .  What  would  6  gross  of  buttons  cost,  at  6  cents  per  dozen? 

Ans.  432  cents. 

62.  AVhat  would  3  great  gross  of  combs  cost,  at  3  cents  apiece? 

Ans.   1 5552  cents. 

63.  What  would  4  score  of  hogs  cost,  at  2  dollars  apiece? 

Ans.    160  dollars. 

64.  A  merchant  purchased  8  fothers  of  iron  ;  how  many  pounds 
was  this?  Ans.    19264. 

6^.  What  cost  2  reams  of  paper,  at  ]j  cents  per  sheet? 

Ans.   1440  cents. 

66.  A  farmer  started  to  market  with  12  dozen  dozen  eggs  ;  he 
broke  a  half  a  dozen  dozen  by  the  way,  and  then  returned  and  got  6 
dozen  dozen  more  ;  how  many  did  he  take  to  market  at  last? 

Ans.  2520. 


174  COMMON    ARITHMETIC.  Scct.  IX 

Article  3.     Fractions  of  Compou"d  Kumbers. 

Case  1.  To  reduce  a  fraction  of  a  higher,  to  a  fraction  of  a  lower 
denomination,  and  fractions  of  lower  to  fractions  of  higher  denomina- 
tions. 


Ex.  1.  Reduce  —  of  a  bushel 
to  the  fraction  of  a  pint. 

Solution. — We  learn  from  Art. 
2,  Obs.  3,  that  we  reduce  integers 
from  larger  to  small  denomina- 
lions  by  multiplication.  Frac- 
tions are  reduced  in  the  same 
manner   as    whole     numbers. — 

Thus:— ^V  X  4  =  yV;  iVX  8 
==  I  ;  I  X  2  =  |.  Ans. 

Or,  by  cancelation  : 

Sect,  yill,  Art.  5,  Obs.  2  and 
3. 


^4 


3 


We  multiply   by 


-4 


4=|-  Ans. 
4,    8  and  2, 
because  4  pecks  make  a  bushel, 
(fee. 

3.  Reduce  y  2  <^^  ^  bushel   to 
the  fraction  of  a  pint. 


5.  Reduce 


of  ahoo'shead, 


Wine  measure,  to  the  fraction  of 
a  P'ill. 


7.  Reduce 


of 


■5-^4  ui  a  barrel, 
wine  measure,  to  the  fraction  of 
a  pint. 

9.  Reduce  -^\^  of  a  barrel, 
beer  measure,  to  the  fraction  of 
a  pint. 

1 1 .  Reduce  ^V  of  ^  French 
ell  to  the  fraction  of  a  nail. 


2.  Reduce  |-  of  a  pint   to  the 
fraction  of  a  bushel. 

Solution. — Integers  are  reduced 
from  a  lower  to  a  higher  denom- 
ination  by   division,      (Art.    2, 
Obs.  3).    Fractions  are  reduced 
in   the    same    manner  as  whole 
numbers.     Thus — |-  -^-  2=|;  §- 
_:_  8  ==  --  •  J-  -1-  4  =  J-  Ans   ' 
Or,  by  cancelation  : 
Sect.  VlII,  Art.  6,  Obs.  2. 
3 
2 
8 

48 


=  xV  Ans. 


We  divide  by  2,  8,  and  4,  be- 
cause 2  pints  make  a  quart,  8 
quarts  make  a  peck,  &c. 


4.  Reduce  ,j  of  a  pint  to  the 
fraction  of  a  bushel. 

6.  Reduce  |  of  a  gill  to  the 
fraction  of  a  hogshead. 

8.  Reduce  \  a  pint  to  the  frac- 
tion of  a  barrel,  wine  measure. 

10.  Reduce  y  of  a  pint,  beer 
measure,  to  the  fraction  of  a  bar- 
rel. 

12.  Reduce  J  of  a  nail  to  the 
fraction  of  a  French  ell.  / 


How  do  we  reduce  fractions  of  higher,  to  fractions  of  lower  denominations? 
How  do  we  reduce  fractions  of  lower,  to  fractions  of  higher  denominations? 


Art.  3. 


REDUCTION. 


175 


13.  A  cucumber  grew  j^/ts- 
of  a  mile  in  leni^th;  what  fraction 
was  that  of  a  foot? 

15.  Reduce  xyTpo  f>f  an  acre 
to  the  fraction  of  a  foot. 

17.  Reduce  jilj^  of  a  solid 
yard  to  the  fraction  of  a  solid 
inch. 

19.  Reduce  ~j  of  a  p  und 
Troy,  to  the  fraction  of  a  pji.ny- 
weii^ht. 

21 
the  fraction  of  an  ounce. 

23.  Rvduc<'  jjj  of  a  pound  to 
the  fraction  of  a  dram. 

25.  Reduce  ylg-  of  a  pound  to 
the  fraction  of  a  scruple. 

27.  Reduce  ^3^20  of  a  day  to 
the  fraction  of  a  minute. 

29.  Reduce  jjj-^o  of  a  circle 
to  the  fraction  of  a  minute. 


Reduce  a^l-oTr  of  a  ton  to 


14.  A  cucumbe*'  gr  w  j  of  a 
foot  in  ien^-:  ;  what  fraction  is 
that  0"  -I  m     ? 

16.  Reduce  y|  of  a  foot  to  the 
fraction  of  an  acr  . 

18.  Reduce  |t  of  a  solid  inch 
to  the  fractioa  of  a  solid  yard. 


20.    Reduce   f   of    a   penny- 
weight to  the  fraction  of  a  pound. 


22.  Reduce  ~~  of  an  ounce  to 
the  fraction  of  a  ton. 

24.  Reduce  |  of  a  dram  to  the 
fracLiaa  oi  a  j  ound. 

26.  Reduct  |-J  of  a  scruple  to 
the  fraction  of  a  pound. 

28.  Reduce  -^  of  a  minute  to 
the  fraction  of  a  day. 

30.  Reduce  J  of  a  minute  to 
the  fraction  of  a  circle. 


Case  2.     To  reduce  fractions  of  co?npound  numbers  to  integers  of 
the  same,  and  also  to  reduce  the  integers  back  to  fractions. 


Ex.  1.  Reduce  ^  a  bushel  to 
pecks. 

Solution. — As  4  peeks  make  a 
bushel,  it  is  evident  that  i^  a  bush- 
el contains  |-  X  4  =  2  pecks, 
Ans. 

3.  Reduce  |  of  a  pint  to  gills. 

5.  Reduce  j  of  a  peck  to 
quarts. 

7.  Reduce  j  of  a  bushel  to 
pecks  and  quarts. 

Solution. — Multiply  |- by  4,  we 
obtain  \"  =  2|-  pecks  ;  multiply- 
ing J  by  8,  we  obtain  ^j  =  4 
qts.  Ans.  2  pks.  4  qts. 


2.  What  part  of  a  bushel  is  2 
pecks? 

Solution. — As  4  pecks  make  a 
bushel,  it  is  evident  that  2  pecks 


are 


i-  a  bushel. 


4.  What  part  of  a  pint  is  3  gills? 

6.  What  part  cf  a  peck  is  2 
quarts? 

8.  Reduce  2  pks.  4  qts.  to  the 
fraction  of  a  bushel. 

Solution. — We  first  reduce  the 
4  qts.  to  the  fraction  of  a  peck  by 
dividing  by  8.  (4-^-8=  i.)  We 
then  have  2^  =  J   pecks,    which 


How  do  we  reduce  fractions  of  compound  numbers  to  integers  of  the  same? 
E.*p4;iin  the  operation  of  Ex.  7  and  8,  Case  2. 


m 


COMMON    ARITHMETIC, 


Sect.  IX 


This  operation  may  be  shown 
thus — 

5  numerator. 

4  pecks  in  a  bushel. 


denom.    8)20 


pks. 


2+4 

8  qts. 


peck. 


denom.  8)32 


4  qts. 

Or,  the  following  form  may  be 
adopted  if  preferred  by  the  stu- 
dent. 

denom.  II  ?><^=^  pks.  &1  rem. 
^1  1X8=4  qts. 

We  cancel  4  into  8,  leaving  2; 
dividing  5  by  2  we  obtain  2  pks. 
and  1  remainder,  which  we  place 
below  the  5.  We  multiply  1  by 
8,  which  cancels  against  the  2 
leaving  4,  and  1X4=4. 

Ans.  2  pks.  4  qts. 

It  will  be  perceived  that  the 
number  at  the  left  is  not  canceled 
unless  it  divides  the  numbers  at 
the  right  without  a  remainder, 
although  we  divide  the  numbers 
at  the  right  by  the  number  at 
the  left. 

This  process  is  the  same  as  in 
Art.  2,  General  Rule  for  Reduc- 
tion Descending,  except  that  we 
divide  by  the  denominator  after 
each  multiplication. 

9.  Reduce  ~q  of  a  hhd.,  wine 
measure,  to  its  proper  value. 

11.  Reduce  |  of  a  bl.,   beer 


we  reduce  to  the  fraction  of  a 
bushel  by  dividing  by  4.  (|-  -^- 
4  =  f.)  This  gives  J  of  a  bush- 
el as  our  answer. 

The  opeiat.on  may  be  shown, 
thus — 


4—  1  — ^ 


^=|=f  of  a  bushel  Ans. 

The  reasoning  is  the  same  as 
above.     Hence — 

To  reduce  integers  of  com- 
pound numbers  to  fractions  of  the 
same,  we  have  this 

Rule  . — I .  Divide  the  lowest  de- 
nomination ly  that  number  which  it 
takes  of  this  to  make  a  unit  of  the 
next  higher  denomination,  and  pre- 
fix to  the  result  the  number  given  of 
this  higher  denomination. 

II, — Divide  this  in  the  same 
manner  as  the  first;  so  continue  to 
do  until  it  is  reduced  to  the  fraction 
re  ui    d 

Remark. — If  any  denomination  is 
wanting,  place  a  cipher  in  its  stead. 


measure,  to  its  proper  value, 
13,  What  is  the  value  of 
a  yard,  cloth  measure? 


of 


'V 


10.  Reduce  18  galls.  3  qts.  1^ 
p!s.  to  the  fraction  of  a  hogs- 
head. 

12,  Reduce  14  galls.  1  qt.  1^ 
pts.  to  the  fraction  of  a  barrel. 

14.  Reduce  2  qrs.  2|  na.  to 
the  fraction  of  a  yard. 


How  do  we  proceed  in  canceling?     To  what  does  this  process   correspond? 
How  do  we  reduce  integers  of  compound  numbers  to  fractions  of  the  same? 


Art.  3. 


FRACTIONS   Of   COMPOUND   JfUMBftRS. 


1T7 


15,  What  is  the  value  of  4  of 
a  mile? 


an  acre: 

19.  What  is  the  value  of  |  of 
u  cord? 

21.  What  is  the  vaiue  of  J  of 
%  pound  Troy? 

23.  What  is  th-e  value  of  — 
X)f  a  ton? 


16.  Reduce  4  fur.  22   rds.   4 
j&s,  2  ft.   ly   in.  to  the  fraction 
I  of  a  mife. 
17.  What  is  the  value  of  J  of  I       18.  Reduce  3  R.  20  sq,  rds.  to 

the  fractron  of  an  acre. 

20.  Reduce  85  sq.  Ft.  576  sq. 
in.  to  the  fraction  of  a  cord. 

22.  Reduce  6  oz^  13  pwt.  8 
grs.  to  the  fraction  of  a  pound. 

24.  Reduce  9  cwt  23  lbs,  1 
oz.  3~  drs.  to  the  fraction  of  a 
•ton, 

26.  Reduce  53.  33.  1©,  18/y 


many 


25.  What  is  the  value  of  y\ 
lof  a  pound,  Apothecaries  weight? 

27.  J  of  a  month  is  how 
days,  hours,  <fec. 

29.  What  is  the  value  of  f  of 
•  *,  circle? 

V  Case  3.  Toreduce  the  decimal 
^df  a  comjmund  wxmber  to  Us  proper 
value. 

Ex.  1.  Reduce.625  of  ahush" 
%1  to  its  proper  value;  that  is,  to 
pecks  and  quarts. 

Operation. 
,625 
4 


grs  to  the  fraction  of  a  pound. 

28.  3  wks,  1  d.  9  hrs.  36  min, 
is  what  fraction  of  a  month? 

30.  Reduce   102^  51'  25f"  to 
j  the  fraction  of  a  circle, 
I      Cass  4. — To  change  numbers 
I  of  different  denominations  to  a  de~ 
j  ci/mal  of  one  denomination. 

I      2.  Reduce  2  pks.  4  qts.  to  the 
decimal  of  a  bushel. 

X)perdiion. 
8  4.0 


pics.  2  .500 


qts.  4.000 
Ans.  2  pks. 


4  qts. 


.  625=yVV^.  (Sect.  YIII,  Art. 
J7,  Obs,  2.)  Reducing  this  as 
m  Case  2,  Ex.  7,  we  obtain  2  pks. 
4  qts.  as  our  answer.  But  since 
iftj^  denominator  of  a  decimal 
miction  is  1  with  ciph.crs  anix3xed 
(Sect.  VIII,  Art,  7,  Obs.  2,) 
we  may  dispense  with  the  divis- 


412.500 


Ans.   .^2S 


As  we  have  before  said,  (Sect. 
Vill",  Art.  7,  Obs.  -9)  ti  whole 
number  can  be  reduced  to  a  deci- 
mal by  annexing  ciphers.  Hence, 
4  qts.=4  .  0  qts.,  and  this  divided 
by  8,  (because  8  qts.,  make  1 
peck,)  gives  ,  5  of  a  peck.  We 
then  have  ^2  . 5  pks.  wh-ich  divi- 


If  any  denomination  is  wanting,  how  do  we  proceed?  How  do  wereducp  the 
decimal  of  a  compouud  namb'er  t*  integers?  How  do  we  reduce  the  iftt«ge4-8 
%'ack  tb  defcimalsT  .--"v„k^, 

^A 


rt9 


COMMON    ARITHMETIC. 


Sect.  IX 


ion,  and  point  off  as  in  multipli- 
cation of  decimals.  (Sect.  YIII, 
Art.  11.  Rule.) 

Remark. — The  learaef  w\\\  pefceire 
that  the  numbers  attlie  left  of  the  deci- 
mal point  are  not  muitiplied,  they  not 
being  decimals,  bul  whole  numbers. — 
Hence — 

To  change  a  decimal  of  a  com- 
pound number  to  its  proper  value, 
we  have  this 

Rule  I. — Multiply  the  giveti 
decimal  by  that  number  which  it 
takes  of  the  next  less  denomination 
to  make  1  of  this  higher,  and  point 
off  a»  in  Midtiplicatwn  of  Deci- 
mals. (Sect.  VIII.  Art.  11, 
Rule.) 

II .  Proceed  in  the  same  manner^ 
midtiplying  the  decimals  only,  of 
each  succeeding  product,  until  no 
decimals  remain,  or  it  is  reduced  to 
the  lowest  denomination  of  the  kind 
to  ichich  it  belongs.  The  numbers 
at  the  left  of  the  diffei^ent  separating 
jpoints,  will  form  the  required  an- 
swer. 

Note. — If  after  any  multiplication 
there  is  no  wl-iole  nKinber  at  tlie  left 
of  the  separatiiag  point,  a  cipher  may 
be  put  in  its  siead. 

3.  Reduce  .  75  of  a  bushel  to 
its  proper  value. 

5,  Reduce  .  25  of  a  hhd.,  wiae 
measure  to  its  proper  value. 

7.  Reduce  .  8  of  a  gallon  to 
its  proper  value. 

9.  Reduee  .  5625  of  a  yard  to 
quarters  and  nails. 

11  Reduce  .8125  of  a  miTe 
to  furlongs  ;  nd  rods. 

13.  Reduce  .5  of  a  yard  to 
•  fest  and  inches. 

15.  Reduce  ,  9375  of  an  acre 
to  its  proper  quantity  ^ 


ded  by  4,  (because  4  pks.  make 
1  bushel,)  gives  .  625  of  a  bush- 
el as  our  answer.     Hence — 

To  change  numbers  of  differ- 
ent denominations  to  a  decimal  of 
one  denomination,  we  have  this 

Rule  I. — Begin  with  the  least 
denomination:  annex  ciphers  to  the 
right  and  divide  it  by  that  number 
vjhich  it  takes  of  this  to  make  a 
unit  of  the  7iext  higher,  and  point 
off  a  decimal  for  every  cipher  an- 
nexed. 

11.  To  the  decimal  thus  obtained 
prefix  the  number  given  {if  any,) 
of  this  higher  denomination,  a7id 
divide  as  before,  pointing  off  as  in 
Division  of  Decimals.  (  Sect. 
VIII,  Art.  12,  Rule,aDd  Rem.  3,) 
So  continue  to  do  until  it  is  reduced 
to  the  denomination  required, 

Note  1. —  If  any  deaonai nation  is 
wanting,  a  cipher  may  be  put  in  its 
place  at  the  left  of  the  decimals. 

2.  Ciphers  must  be  prefixed  to  the 
several  quotients,  (not  to  the  integral 
parts,)  when  it  is  necessary,  to  co-m- 
plete  the  decimal  places., 


4,  Redace  3  pks.  to  the  deci- 
mal of  a  bushel, 

6,  Reduce  15  g^Vm.  3  qts.  to 
the  decimal  of  a  hhd.,  wine  mea- 
sure- 
s'. Redn^de  5  qts.  1.6  gi.  to 
the  decimal  of  a  gallon. 

10.  Reduce  2  qrs,  1  na.  to  th© 
decimal  of  a  yard.  ^(tf 

n.  P>educ«;  6  fur.  20  rds.  to 
the  decimal  of  a  mile. 

14.  Reduce  1  ft.  6  in.  to  the 
decimal  of  a  yard. 

16.  Reduce  3  R.  30  sq,  rds.  to- 
tk^  decimal  of  an  aci^e.  . 


Art.  4. 


ADDITION   OP   COMfOUNO   NlTMBERS. 


17 


17.  Reduce  .  73 125  of  a  pound 

Troy,  to  its  proper  value. 

19.  Reduce  .7896875  of  a 
ton  t .)  Its  proper  value. 

21.  Reduce  .3984375  of  a 
pound  Apothecaries'  weight,  to 
its  proper  value. 

23.  Change  .  5625  of  a  year 
to  months  and  weeks. 


18.  Reduce  8 oz.,  15pwts.,  1 
grs,  to  the  decimal  of  a  pound. 

20.  Reduce  15  cwt.   3  grs.  4 
lbs.  6  oz.,  to  the  decimal  ot  a  ton. 

22.  Reduce  43.  63.  15  grs.  to 
the  decimal  Of  a  pound. 

24.  Reduce  6  mos.  3  wks.^  to 
the  decimal  of  a  year. 


RiE^tAftS;. — There  are  two  othercases  of  Reduction  of  Decimals  of  Compound 
numbers  but  they  are  not  deemed  of  suiHcieut  importance  to  present  them  in 
this  work.  We  give  below,  an  example  of  each,  with  their  solution,  by  the 
aid  of  which  the  observing  pupil  can  perceive  their  application  and  use. 


1 .  Reduce  .  5  of  a  pint  to  the  |      2 
decimal  of  a  gallon. 

Operation. 


2 


50. 


Reduce  .0625  of  a  gallon 
to  the  decimal  of  a  pint. 
Operation^ 
.0625 


2500 


Ans.  .0625. 
We  annex  ciphers  to  our  deci- 
mal, which  does  not  alter  its  val- 
ue, (Sect.  VIII,  Art.  I,0bs7.  6.) 
and  divide  by  2  and  4  because  2 
pts.  make  a  quart,  &c.  We  point 
off  as  in  Division  of  Decimals. — 
(Sect.  YIII,  Art.  12.  Rule.) 


.  2500 
2 


Ans.  .5000 
We  multiply  by  4  and  2,  be- 
cause 4  qts.  make  a  gallon,  <fec., 
and  point  off  as  in  Multiplication 
of  Decimals.  (Sect.  VIIl,  Art, 
11,  Rule.) 


Article  4.     Addition  of  Compound  Numbers. 


Ex.  1 .  There  are  two  bushels  of  wheat  in  one  bag,  one  bushel  in 
another,  and  2  bu.,  3  pks,  in  another.     How  much  in  all? 

Ans.  5  bu.  3  pks. 

2.  Thomas  has  1  lb.  4  oz.  of  raisins,  James  has  2  lbs.  6  oz.,  and 
.^Wliam  has  2  lbs.  5  oz.     How  many  pounds  have  they  all? 

■IP^  Ans.  5  lbs.  15  oz. 

3.  A  man  sold  at  one  time  18  bu.  3  pks.  6  qts.of  oats,  at  another 
lime  23  bu>  2  pks.  4  qts.,  and  at  another  time  12  bu.  I  pk.  7  (jts. 
How  much  did  he  sell  in  ill? 


I8t)  COMMON    ARltHMfirfC.  S6ct.    IX 

Operation,  We  first  write  down  the  numbers,  placing 

bu,  pks.  qts.  those  of  the  same  kind  under  each  other.— ■ 
18_-3__6  "We  commence  at  the  right  hand,  or  lowest 
23  __  2  __  4  denomination,  and  add  it  as  in  simple  nuni- 
12  ..   I   _-   t       bers;  7  +  4+6  =  17qts.   =  2pks.  Iqt.; 

setting  down  the  1  qt.,  we  carry  the  2  pks. 

Ans.  55__  0  __  1  to  the  next  column;  1 -|-2 -(- 3  =  6  pks. 
and  2  pks  to  carry  make  8  pks.;  8  pks  =  2  bu.  and  no  remainder; 
setting  down  a  cipher  in  the  place  of  pecks,  (because  we  had  no 
pecks  remaining.)  we  carry  the  2  bu.  to  the  column  of  bushels,  and 
add  as  in  simple  numbers,  setting  down  the  entire  sum. 
From  this  illustration  we  derive  the  following 

RULE  FOR  THE  ADDITION  OF  COMPOUND  NUMBERS. 

I.  Set  down  the  nu7nhers,  placing  those  of  the  same  denomination 
under  each  other.     (Sect.  IT,  Art,  2,  Obs.  4.) 

II.  Begin  with  the  lowest  denomination  ;  add  together  the  numbers 
given  of  this  denomination,  as  in  Addition  of  Simple  numbers,  and 
divide  their  sum  by  that  number  which  it  takes  qf  this  denomination  to 
make  a  unit  of  the  next  higher  ;  place  the  remainder  under  the  column 
added,  and  carry  the  quotient  to  the  next  denomination, 

III.  Add  the  remaining  denominations  in  the  same  iiianner  as  the 
first. 

Proof. — 2 he  same  as  in  Addition  of  Simple  numbers^  (Sect.  Il, 
Art.  2  Obs.  6.) 

Remakk  1. — If  any  denomination  is  wanting,  supply  its  place  with  a  ciphei*. 
2.  The  principles  of  Addition  of  componnd  and   simple  numbers  are  the 
same.     (Art.  1,  Obs.  6.)     This  can  be  shown  by  the  following  example: 

Add  together  2436,  1589,  1612,  and  2849. 

Operation.       .  94-2-j-9+6  =  26;26-M0  =  2 

thou.  hund.  tens,  units,  and  6  remainder.     4-f-l-(-8-f^3=£ 

2  _.   4  __   3  ._   6  1 6  and  2  to  carry  =  18  ;   18 -MO  =  1 

1  _  _  5  _  _  8  _  _  9  and  8  rem.  8  +  6  +  5  +4  =  23  and 
1__6_-1__2  Ito  carry  =  24,  24  -i-  10  =  2   and  4 

2  __  8  ._  4  -_  9  rem.  2  +  1  -f  1  -[- 2  =  6  and  2  to 
carry  =  8.                                      ^H|^ 

Ans.  8  __   4  __   8  __   6  ^ 

What  is  the  rule  foi*  addition  of  compound  numbers?  What  is  the  proof? 
If  any  denomination  is  wanting,  how  do  we  proceed?  What  is  the  diffefence 
between  the  principles  of  addition  of  simple  and  compound  numbers? 


'Art.    4.  ADDITION    OP    COMPOUND    NUMBERS.  tSl 

It  will  be  perceived  that  we  divide,  &c.,  the  same  as  in  tlie  above 
example,  but  as  the  remainder  is  alvvays  the  right  hand  figure,  this 
formal  process  of  dividing  may  be  dispensed  with. 

4.  A  merchant  bought  at  one  time  36  bu.  3  pks.  6  qts  of  wheat ; 
at  another  time  25  bu.  2  pks.  3  qts.;  at  another  time  30  bii.  1  pk.  7 
qts.;  at  anothertime  28  bu.  4  qts.;  and  at  another  time  17bu.  2  pks.; 
how  many  did  he  buy  in  all?  Ans.  I'i8  bu.  2  pks.  4  qts. 

B.  Bought  5  loads  of  oats;  the  first  contained  27  bu.  8  qts.;  the 
second,  37  bu.;  the  third,  26|  bu,  6\  qts.;  the  fourth,  36^  bu.  1 
pk.  6  qts.;  and  the  fifth,  35  bu.  3 J  pks.;  how  many  bushels  in  all? 

Ans.  164  bu.  3^  qts. 

6.  A  wine  merchant  has  1  hhd.  27  galls.  3  qts.  of  Madeira ;  2 
hhd.  30  galls.  3^  qts  of  Port;  61/o-galJs.  Champaign,  and  3  hhds. 
If-  bis.  of  Claret  wine  ;  besides  these  he  has  2 J  hhds.  of  Brandy  ; 
and  1  hhd.  24  galls.  2|  qts.  of  other  liquors.  How  much  has  he  in 
all?  Ans.   12  hhds.  1  bl.  31  galls.  2  qts.  1  pt.  3|  gi. 


.     Note. — The  learner  will  bear  in  mind  that  the  remainder,  after  dividing  by 
*3li  is  half  gaHotis,  not  whole  gallons. 


7.  A  wine  merchant  sold  at  one  time  45  galls.  3  qts.  1  pt.  3  gi.  of 
wine;  at  another  time,  36  galls.  2  qts.  1  gi.;  at  another  time  49  galls 
1  pt.;  at  another  time  57  galls.  3  gi.;  and  at  another  time  38  galls.; 
how  much  did  he  sell  itl  all? 

Ans.  3  hhds.  37  galls.  2^qts.  1  pt.  3  gi. -"J 

8.  A  brewer  sold  6  casks  of  beer:  the  first  contained  33  galls.  2 
qts.  1  pt.;  the  second  28  galls.  1  pt.;  the  thirJ,  34j  galls.;  the  fourth, 
26  galls.  3^  qts.;  the  fifth,  18  galls.  2  qts.;  how  much  was  there  in 
all?  "  Ans.  2  hhds.  33  galls.  3  qts.  1  pt. 

9.  Bought  4  pieces  of  cloth  ;  the  first  contained  22  yds.  3  qrs.  2 
na.;  the  second,  21 J  yds.;  the  third,  18  yds.  2 J  qrs.;  the  fourth,  33 
yds.  2  qrs.  2jna.;  how  many  yards  in  alU  Ans.  97. 

10.  If  it  takes  3  yds.  3j  qrs.  of  cloth  to  make  a  coat,  2  yds.  2  qrs. 
3  na.  to  make  a  pair  of  pants,  and  2|  qrs.  to  liiake  a  vest  how  many 
yards  will  it  take  for  the  whole?  Ans.  7  yds.  1  qr.  1  na. 

11.  Thomas  walked  due  north  43  miles,  7  fur.  29  rds.  12  ft.,  and 
James  walked  due  sQuth  bQ  miles,  5  fur.  34  rds.  14  ft.; how  far  apart 
ai^giey?  '    '"    '  Atis.   100  M.  5  fur,  24 rds.  9^  ft. 

T2.  A  gentleman  traveling,  rode  the  first  week  167  M.  7  fur. 
38  rds.;  the  tecond  week,  180  M.  6  fur.  27  rds.;  the  third  week, 
173  M,  5  fur.  36  rds.;  and  the  fourth  week  77  M.  3  fur.  19  rds.; 
hovx  ^any  miles  did  lie  travel?  Ans.  600. 

"""!"■ "  '     -^  e:..    '-a.;      ij   ^  ■    .: "  '      '     ^ — " '■ T^- 

Show  by  tlie  example "giveii  why  this  is  the  case?  Why  do  we  not  divide  in 
this  mannerm  addition  of  simple  nunib'ers? 


182  COMMON  ARtTiiMEttc.  Sect,  tX 

13.  A  man  has  4  farms;  the  first  contains  120  A.  3  R.  27  sq^ 
rds.;  the  second,  112  A,  2|  R.;  the  third,  94j  A.;  and  the  fourth> 
87yV  ^l  ^ow  many  acres  does  he  own  in  all? 

Ans.  416  A.  1  R.  9  sq.  rds.  or  perches. 

14.  There  are  6  piles  of  wood;  the  first  contains  3  C.  64  sq.  ft. 
1126  s.  in.;  ihe  second,  4  C.  112  s.  ft.  1692  s.  in.;  the  third,  5  C. 
127  s.  ft..  1724  s.  in.;  the  fourth,  5|  C;  the  fifth,  6  C.  109|  sq.ft.; 
and  the  sijcth  contains  3  C.  123  s.  ft.  498  s.  in,;  how  many  cords  in 
the  whole?  '  Ans.  30  C.  7  c.  ft. 

15.  A  silversmith  purchased  4  bars  of  silver:  the  first  weighed  2 
lbs.,  9  og.  15pwts.  22  grs  ;  the  second,  1~  lbs.;  the  third,  3  lbs.  1| 
oz.;  and  the  fourth,  2  lbs.  11  oz.  9  grs.;  how  much  did  they  all 
weiiih?  Ans.   10  lbs.  8  oz.  4  pwts.  S3  grs. 

16.  Sold  atone  time  1  cwt.  2  qrs.  23  lbs.  14^^  oz.  of  sugar;  at 
another  time  2  ~  cwt.;  at  another  time  2  cwt.  1  qr.  16|  lbs.;  and 
at  another  time  2  cwt.  3  qrs.  19  lbs.  13  oz.  8  drs.;  how  much  was 
sold  in  all?  Ans.  9  cwt.  3  qrs.  23  lbs. 

17.  Bought  4  stacks  of  hay;  the  first  coniained  3  ^T.  19  cwt.  2 
qrs.  15  lbs.;  the  second,  2  T.  14  cwt.  3i  qrs.;  the  third,  3y\  T.;  and 
the  fourth,  4  T.  19y  cwt.,  how  many  tons  in  all? 

Ans.  15. 

18.  An  Apothecary  mixed  one  compound  with  5  simples:  the 
first  weighed  1  lb.  23.  43.  1  9.;  the  second  weighed  9  3-  7  3-  ^  3- 
16  grs.;  the  third,  6  5.  53.  19  grs.;  the  fourth,  3  3.  15  grs.,  and 
the  fifth,  73.   19.;  what  was  the  weight  of  the  whole? 

Ans.  2  ft).   113.   13.   10  grs. 

19.  A  gentleman  attended  school  until  he  was  21  yrs.  212  da.  of 
age;  he  then  traveled  15  yrs.  84  da.;  he  then  married  and  lived  wilk 
his  wife  14  yrs.  176  da.,  when  his  wife  died:  after  this  he  hved  23 
yrs.  219  da.;  what  was  his  age  when  he  died? 

Ans.  76  yrs.  240  da.  12  hrs. 

20.  Upon  what  day  of  the  year  does  Christmas  come  in  leap  year? 

Ans*  the  360th. 

21.  On  what  day  of  a  common  year  does  the  fourth  oi  July  hap- 
pen? Ans.  the  185th. 

22.  Boston  is  71°  3'  west  longitude  from  Greenwich,  Washing- 
ton 6°  40'  farther  west,  Cincinnati  6°  44'  still  farther,  and  theAet* 
ern  line  of  Vancouvers'  Island  38°  33'  still  farther  west.  Re- 
quired the  longitude  of  Vancouvers'  Island  from  Greenwich? 

Ans.   123°. 

23.  Bouglit  4  casks  of  butter:  tlic  first  contained  1  fir,  29  lbs.; 
the  second,  2  fir.  43  lbs.;  the  third,  1|  fir.;  jmd  the  fourth,  2j  Jr.; 
hov  much  was  thei*e  in  all?  '  '  Ami.  8  gr.  f  f ' 


Art..^  SUBTRACTION   OF   COMPOUSD  NUMBERS.  1S3 

24.  Bought  G  boxes  of  eggs:  the  first  contained  2  dozen  dozen; 
the  second.  7  dozen  dozen;  the  third,  12  dozen  dozen;  the  fourth, 
half  a  dozen  dozen;  the  fifth,  1  great  gross;  and  the  sixth,  12 
score.     How*  many  eggs  in  all,  and  how  many  dozen? 

Ans,   6064  eggs=  4S2  dozen. 

25.  Bought  1  ream,  18  quire,  9  sheets  of  paper  at  one  time;  2 
reams,  10  sheets  at  another  time;  1|- r  ams  at  another  time ;  2| 
reams  at  another  time;  and  1  ream,  8  quire,  13  sheets  at  another 
time;  how  much  did  1  buy  in  all?  Ans.   10  reams. 

26.  To  J  of  a  bushel,  add  \  a  peck.  Ans.  3  pks.  4  qts. 

27.  To  |-  of  a  hhd.,  wine  measure,  add  J  of  a  barrel, 

Ans.   1  hhd.  15  galls.  3  qts. 

Note. — Reduce  both  to  their  value  in  iuteffers,  and  than  add  them. 

28.  To  J  of  a  mile,  add  f  of  a  rod. 

Ans,  6  fur.  27  rds.   5  ft,  6  in. 

29.  To  J  of  a  cwt.,  add  |of  a  pound. 

Ans.  3  qr.  13  lbs.  4  oz, 

30.  To  i  a  year,  add  |  of  a  day.  Ans.   1 83  da.  7  hr. 

Article  6.     Subtraction  of  Compound  Numbers. 

Ex.  1.  There  r.ere  2  bu.  3  pks  of  oats  in  a  box,  and  1  bu<  2 
pks.  were  taken  out;  how  much  was  left?  Ans.   1  bu.  I  pk. 

2,  From  a  box  containing  8  bu.  2  pks,  6  qts.,  5  bu.  1  pk,  2  qts. 
were  taken,  how  much  remained?  Ans.  3bu.  1  pk,  4  qts. 

3.  A  man  having  64  bu.  1  pk.  3  qts,  of  wheat,  sold  45  bu.  3  pks. 
6  qts,;  how  much  had  he  left?  Ans,  18  bu,  1  pk.  5  qts. 

Operation,  We  cannot  take  6  qts.  from  3  qts.,  but  we 

bu.  pks.  qts,  can  borrow  1  pk.  (8  qts.)  from  the  next  or- 
64  --  1  __  3  der;  then  8  qts.  -I-  3  qts,  =  11  qts.,  11  qts. 
45  __  3   __    6       — 6  qts.  =  5  qts.;  we  set  down   the  5   qts., 

and  as   we    borrowed    1  pk.  we  must  carry 

Ans.  18__1_..  5  Ito  the  column  of  pecks,  to  make  up  for 
that  we  borrowed  14-3=4  pks.;  we  cannot  take  4  pks.  from  1  pk., 
but  we  can  borrow  1  bu.  (4  pk-.)  from  the  column  of  bushels:  4 
pks. -(-  1  pk.  =  5  pks. — 4  pks.  =  1  pk.  We  set  down  the  1  pk,,, 
and  carry  1  (to  pay  for  that  Ave  borrowed)  to  5  making  6;  6  from  14 
=  8,  &c. 

This  process  may  also  be  shown,  thus — 

bu.     pks.  qts,         bu.     pks,  qts.         This  operation  is  so  simple 
64._1__3    =    63_-4_.ll     that  it  requires  no  explana- 
45  .-   3._   6     tion, 

Ans.    18  ..    1  ..   6 


184  coMJyiON  ARi^iiMEtic.  S'ect.  IX 

From  these  illustrations  we  derive  the  following 

RULE  FOR  THE  SUBTRACTION  OF  COMPOUND  NUMBERS. 

I.  Write  the  numbers,  the  hss  under  the  greater,  placing  those  of  the 
same  denomination  under  each  other. 

II.  Commence  with  the  lowest  denomination:  take  successively  each 
denomination  from  the  one  above  it,  placing  the  result  below  as  in  sim- 
ple numbers.     (Sect,  II [,  General  Eule.) 

III.  If  ike  lower  number  should  be  the  lar^d,  add  to  the  upper 
number  a  unit  of  the  next  higher  denomination,  after  which,  subtract  as 
usual,  mid  carry  1  to  the  next  dx^nominatlon  for  that  which  was 
hortowed. 

Proof, — The  same  as  in  Subtraction  of  Simple  Numbers.  (Sect» 
III,  Art.  2,  Obs.  7.)  ^ 

Remark  1.     If  any  denomination  is  wanting,  supply  its  place  with  a  cipher* 
^.  The  principles  of  subtraction  of  si/nple  and  compound  numbers  are  the 
same.     (Art,  1,  Obs.  6.)     This  may  be  shown  by  the  following  examplei 

From  356  take  289, 

\st,   Operation,                                 2d    Operation.  » 
hund.  tens,  units.                          hunds,  tens,  units. 

3  _-   6  __   6  356  =  2  ._  14  __  16  . 

2  ..    8  _-   9                                   2  __     8  __     9  ^^ 


■       "  6   __   7  6  __    7  =  67. 

It  will  be  perceived  that  the  principle  is  the  sanae  in  this  example 
as  in  the  one  above. 

4.  A  farmer  raised  1000  bushels  of  wheat,  and  sold  679  bu,  3 
pks.  6  qts.;  how  much  had  he  left?  Ans.  320  bu.  2  qts, 

5.  A  wine  merchant  sold  from  a  hogshead  of  Madeira,  35  galls. 
3  qts.  I  pt.  2  gi.;  he  then  put  in  28  galls,  1  qt.  3  gi.,  and  afterAvards 
sold  43  galls.  3  gi, ;  hoAv  much  had  he  left? 

Ans.   12  galls.  1  qt,  2  gi, 

6.  A  wine  merchant  drew  4  bis.  I  qt.  1  gi.  of  wine  from  a  cask 
containing  3  bis.  31  galls.  3  qts,  1  pt.  3  gi.  Required — the  quan- 
tity left  in  the  cask.  Ans.   1  pt.  2  gi, 

I  ■   7.  From  If  hhds.  of  Port  wine  was  taken  54|-  galls.     How  much 
was  left?  Ans,  31  galls,  3  qts.  1  pt. 

What  is  the  rule,  for  the  subtraction  of  compound  numbers?  Wliut  is  the 
proof?  If  any  denomiiiatiou  is  vviuitiug,  how  do  we  proceed?  What  is  tlie 
difference  between  the  principles  of  subtraction  of  simple  and  compound  num- 
-bers?     Explain  why  this  is  the  case. 


Art.    5.  SUBTRACTION   OF    COMPOUND   NUMBERS.  185 

8.  From  2|  bbls.  of  beer,  was  taken  34j  galls.  How  mucli  re" 
mained?  Ans.   1  bl.  19  gals.  1  pt. 

9.  From  a  piece  of  cloth  containing  18-  yds.,  was  made  a  coat 
containing  2  yds.  1  qr,  3  na.,  a  vest  containing  2\  qrs,,  and  a  pair 
of  pants  containing  1  ]■  yds.;  how  much  was  left? 

Ans.  13  yds.  3  qrs.  1  na. 

10.  A  traveled  140  miles:  B  followed  after  him  76  M.  7  fur.  35 
rds.;  how  far  apart  were  they?  Ans.  63  M,  5  rds, 

11.  It  is  24  miles  from  Columbus,  0,  to  Delaware;  if  a  man 
travel  15|  miles,  from  Columbus,  how  much  farther  must  he  go  to 
get  to  Delaware?  Ans.  8  M,  2  fur,  26  rds.  11  ft. 

12.  A  man  owned  a  farm  containing  134  A.  3  J  R.;  he  sold  at  one 
time  43  A.  2  R.  27  sq.  rds.,  and  at  another  time  32|-  A.;  how  much 
had  he  left?  Ans.  58  A.  2  R.  23  sq.  rds. 

13.  A  certain  pile  of  wood  contains  17  cord,  112  s.  feet.  1438  s. 
in.;  another  pile  contains  14  C.  123  s.  ft,  1649  s.  in.;  how  much 
more  in  one  pile  than  in  the  other? 

Ans.  2  C.   116  s,  tt.  1517  s.  in. 

14.  A  silversmith  had  a  piece  of  silver  weighing  4  lbs.  8  oz.  16| 
pwts.  from  this  he  made  a  set  of  spoons  weighing  2  lbs,  6  J  oz.,  and 
thimbles  enough  to  weigh  l|-*lbs.;  how  much  was  left? 

Ans.  9  oz.  10  pwts.  20  grs. 

15.  A  box  of  goods  weighed  62  lbs.  12  oz.;  the  box  alone  weigh- 
ed 18  lbs.  14j  oz.;  how  much  did  the  goods  weigh? 

Ans.  43  lbs.  13  oz.  4  drs. 

16.  If  a  waggon  loaded  with  hay  weighs  28  cwt,  3  J  qrs.,  and  the 
waggon  weighs  7  cwt,  3  qrs.  20  lbs.,  what  is  the  weight  of  the  hay? 

Ans.  21  cwt.  =  1  T.  1  cwt. 

17.  A  merchant  having  I2|  cwt.  of  sugar,  sold  at  one  time  3  cwt. 
3  qrs.  21  lbs.,  at"  another  time  2  cwt.  1  qr.  15  lbs.,  and  at  another 
time  62V  cwt.;  how  much  had  he  left?  Ans.  None. 

18.  From  a  package  of  medicine  containing  4  ft).  IO3.  63.  19. 
15  grs.,  was  taken  2  ft).  11  3. 7  3.  29.  18  grs.;  how  much  was  left.? 

Ans.   1  ft).  10  3.  6  3.  1  9.  17  grs. 

19.  A.  worked  at  a  certain  place  2  yrs.  186  da.  12  hrs.;  B.  work- 
ed at  the  same  place  1  yr.  310  da.  14  hrs.;  how  much  longer  did  A. 
work  than  B.?  Ans.  24 li^  da. 

20.  Henry's  age  is  17  yrs.  6  mo.  23  d.;  George's  age  is  10  yrs., 
7  mo.  21  d.;  how  much  older  is  Henry  than  George? 

Ans.  6  yrs.  11  mo.  2  d. 

27,  Washington  was  born  Feb.  2d,  1732;  he  died  Dec.  14th, 
1799;  what  was  his  age  at  the  time  of  his  death? 

Ans.  67  yrs.  9  mo.  22  d. 


180  COMMON    ARtTHMETIO.  ScCt.    IX 

Operation.  We  set  down  the  year,  the  number  of  the 

yrs.     mo.     d.  month,  (calHng  Jan.   1,  Feb.  2,  &c.)  and  the 

1799__12__14  day  of  the  month,  and   subtract    as   usual, 

1732__    2__22  reckoning  30   days  to   the    month,   and    12 

months  to  the  year.     In  this  example,    Dec. 

Ans.     67 ._    9__22  is  the   12th  month,  Feb.    the  2nd.,    (fee- 
Hence — 

Obs.  1.  The  period  of  timehetiveen  any  two  dates  maybe  found 
hy  subtracting  the  earlier  date  from  the  latter,  the  mnntl «  heing  reckoned 
30  days.  This  rule  is  much  used  in  calculating  interest.  (Sect. 
XIV,  Art.  5.) 

21.  A  note  was  given  Jan.  24th,  1841,  and  paid  Aua;'.  12th  1843; 
what  time  did  it  run?  Ans.  2  yrs.  6  mo.  18  d. 

Obs.  2.  The  time  which  elapses  from  the  giving  of  a  note  until  the 
pcyment  of  it  is  called  the  time  it  has  to  run. 

22.  A  man  left  home  Sept.  28th,  1837,  arid  returned  June  18th, 
1840;  how  long  was  he  absent?  Ans.  2  yrs.  8  mo.  20  d. 

23.  How  long  has  a  note  to  run,  which  is  giv.n  Nov.  16th,  1828, 
and  paid  Jan.  3d.  1847?  Ans.   18  yrs.  1  mo.  17  d. 

24.  London  is  51°  32',  and  Gibralter  36°  6' 30"  north  latitude, 
what  is  the  difference  of  latitude  between  the  two  places? 

Ans.   15°  25' 30". 

25.  From  an  area  of  180°,  take  57°  48'  39". 

Ans.   122°  11'  21". 

26.  The  longitude  of  Boston  is  71°  3',  and  that  of  the  Sandwich 
Islands  155°  west  from  Greenwich;  what  is  their  difference  in  lon- 
gitude? ^  Ans.   83°  57'. 

27.  A  man  having  6  dozen  dozen  eggs,  broke  a  half  a  dozen  doz- 
en; how  many  had  he  left?  Ans.  792. 

28.  Take  149  lbs.  from  a  barrel  of  flour,  and  how  much  remains? 

Ans.  47  lbs. 

29.  From  a  barrel  of  pork,  take  97  lbs.  Ans.   103  lbs. 

30.  From  g  of  a  bushel,  take  |  of  a  peck.      Ans.   1  pk.  6  qts. 

31.  From  |  of  a  hogshead,  wine  measure,  take  §■  of  a  gallon. 

Ans.  52  galls*  1  pt* 

32.  From  \  a  mile  take  y^  of  a  furlong.     Ans.  3  fur.  12  rds. 

33.  From  J  of  a  cwt.,  take  —  of  a  quarter. 

Ans.  2  qrs.  21  lbs.  4  oz. 

34.  From  |  of  a  year,  take  J  of  a  day. 

Ans.  243  days  8  hrs. 


35.  From  y-g-  of  a  circle,  take  J  of  a  degree. 


Ans.   156°  37' 30" 


How  is  the  period  of  time  between  two  dates  found?     What  is  meant  by  the 
time  a  note  has  to  run? 


Art.  G.    MULTIPLICATION  AND    DIVISION    OP   COMPOUND  No's.        187 


Article  6.  Multiplication  and 
Ex.  1.  A  farmer  has  6  bags 
of  wheat  eacli  containing  2  bu.  3 
pks.  6  qts.  how  much  wheat  in 
all  the  bags? 

Operation,  * 
bu.     pks.     qts. 
2  .>  3  ._  6 
6 


Division  of  Compound  Numbers* 
j      2.  A  farmer  has  17  bu.  2  pks. 
i  4  qts.  of  wheat  put  up  in  6  bags, 
i  each  containing  an  equal  quantity; 
how  much  is  there  in  each? 
Operation. 
bu.     pks.     qts. 
6)17  _.  2  ..   4 


Ans.    17   ..   2  _-   4 

We  writ  *  our  numbers  as  usu- 
al, placing  our  multiplier  uyder 
the  lowest  denomination  of  the 
mukiplicand,  (qts.)  Then,  6  X 
6  =  36  qts.;  36  qte.  =  4  pks.  4 
qts.;  we  set  down  the  4  qts.  and 
carry  the  4  pks.,  3X6  =  18  pks. 
and  4  pks.  to  carry  make  22  pks.; 
22pks.  =  5bu.  2  pks.;  we  set 
down  the  2  pks.  and  carry  the  5 
bu.;  2  X  6  =  12  and  5  to  carry 
=  17bu. 

From  this  illustration  we  de- 
rive the  following 

RULE     FOR    THE     MULTICATION    OF 
COMPOUND    NUMBERS. 

I.  Write  the  multiplier  under 
the  lowest  denomination  of  the  mul- 
tiplicand. 

II.  Commence  at  the  rigid  hand 
multiply  as  in  simple  numbers  and 
carry  as  in  addition  of  compound 
numbers. 

Proof. — The  same  as  in  sim- 
ple numbers.  (Sect.  IV,  Art  2, 
Obs.  7.) 

Remark  1.  If  any  denomination 
is  wanting  in  th*^  multiplicand,  supply 
its  place  with  a  cipher. 


Ans.  2   _.   3  ...  6 

17-f-6  =  2,  and  5  remainder. 
This  5  is  bushels,  because  the 
dividend,  17,  is  bushels.  (Sect. 
V,  Art.  2,  Obs.  12,  a.)  5  bu.  = 
20  pks.;  20  pks.  +  2  pks.=  22 
pks.;  22  pks.  -i-  6  =  3  pks.  and  4 
pks.  remainder.  4  pks.  =  32  qts. 
32  qts.  +  4  qts.  =  36  qts.;  36 
qts.  -f-  6  =  6  qts. 

From  this  illustration  we  de- 
rive the  following 

RULE    FOR     THE    DIVISION  OF    COM- 
POUND   NUMBERS. 

I.  Write  the  dividend  and  divi- 
sor as  usual.     (Sect.  Y,  Rule.) 

II.  Commence  with  the  highest 
denomination;  divide,  and  place 
the  quotient  as  in  division  of  sim- 
ple numbers.     (J^ect.  Y,  Rule.) 

III.  If  a  remainder  occurs  re- 
duce it  to  the  next  lower  denomina- 
tion and  add  to  it  the  number 
given,  [if  any)  of  this  denom- 
ination; after  which  divivde  as 
usual. 

Proof. — The  same  as  in  Sim- 
ple Xumhers.  (Sect.  Y,  Art.  2 
Obs.  9. 

Remark   3.      If  any  denomination 


What  is  the  rule  for  the  multiplication  of  compound  numbers?  What  is 
the  proof?  If  any  denomiuation  is  wanting  in  the  multiplicand,  how  do  we 
proceed?  What  is  the  difference  between  the  principles  of  the  multiplication 
of  simple  and  compound  numbers?  Explain  by  the  example  Sfiven,  why  this 
is  the  case?  Why  do  we  not  reduce  in  this  way  in  simple  numbers?  V>  hat  is 
tlie  rule  for  the  division  of  compound  numbers?     What  is  the  proof? 


188 


COMMON    ARITHMETIC. 


Sect.  IX 


2.  The  principles  of  multiplication 
of  simple  and  compound  numbers  are 
the  same  as  can  bo  shown  by  the  fol- 
lowing example. 


Multiply  256  by  8, 
Operation, 
thou,  liunds.  tens,  units. 
2  ._  5  ..  6 
8 


Ans.  2  _.  0  __  4  _.  8 
6X8  =  48 -^10  =  4  tens,  8 
units.  5  X  8  =  40,  and  4  to 
carry  =44;  44  -^  10  =  4  hund. 
4  tens.  2X8  =  16  and  4  to 
carry  =  20  ;  20-r-  10  =  2  thou. 

0  hun. 

It  will  be  perceived  that  there 
is  no  difference  between  this  op- 
eration, and  that  of  Ex.  1. — 
But  as  the  number  that  we  set 
down  is  always  the  right  hand 
figure  of  the  partial  product,  it 
is  evident  that  the  above  formal 
process  is  unnecessary. 

3.  A  farmer  took  7  loads  of 
wheat  to  market,  each  load  con- 
taining 22  bu.  2  pks.  5  qts.;  how 
much  was  there  in  all? 

5.  A  wine  merchant  has  9 
casks  of  wine,  each  containing  20 
galls.  1  pt.;  how  much  had  he  in 
all? 

7.  A  gentleman  has  8  casks  of 
brandy,  each  containing  45  galls. 

1  pt.;  how  many  gallons  has  he 
in  all? 

9.    How  many   gallons    in    5 


in  the  divdend  is   wanting  supply    its 
place  with  a  cipher. 

4.  The  principles  of  the  division  of 
simple  and  compound  numbers  are  the 
same,  as  can  be  shown  by  the  follow- 
ing example: 

Divide  2048  by  8. 
Operation. 
thou.  hund.  tens,  units. 
8)2  __  0  _-   4  ._   8 


Ans.  2  _.  5  _.  6 
2  thou.  X  10  =  20  hund.;  20 
-^-  8  =  2  hund.  and  4  hunds. 
rem.:  4  X  10  =  40  tens,  and  4 
tens  =  44  tens*  44  -r-  8  =  5  tens 
and  4  tens  rem.;  4  X  10  =  40 
units,  and  8  units  =48  units  ; 
48-^8  =6  units. 

It  will  be  perceived  that  the  op- 
eration is  similar  to  that  of  Ex. 
2.  But  the  denomination  given 
only  occupies  the  pi  ace  of  the 
cipher,  when  added  to  the  one 
that  is  reduced,  this  process  may 
be  dispense:!  with.  (Sect.  V, 
Art.  2,  Ex.  9,  and  Obs,  13.) 

4»  If  7  equal  loads  of  wheat 
contain  158  bu*  2  pks.  3  qts*, 
how  much  does  each  contain? 

6.  If  9  casks  of  wine  contain 
181  galls.  1  pt.,  how  many  gall- 
ons does  each  contain,  each  con- 
taining an  equal  quantity? 

8*  If  8  casks  of  brandy,  con- 
taining each  an  equal  quantity, 
hold  361  galls.,  how  many  gal- 
lons are  there  in  each? 

10*  If  5  equal  sized  flasks  con- 


If  any  denomination  is  wanting  in  the  dividend,  how  do  we  proceed? — 
What  is  the  difference  between  the  principles  of  the  division  of  simple  and 
compound  numbers?  Explain  why  this  is  tlie  case.  Why  is  not  this  process 
adopted  in  simple  numbers? 


Art.^.    MULTIPLICATION  AND  DIVISION  OF  COMPOUND  NO.'s  189 


flasks  of  beer,  each  containing  6 
galls.  2  qts.  1  pt? 

1 1 .  How  many  yards  of  cloth 
will  it  take  to  make  7  coats,  each 
containing  3  yds.  2  qrs.  1  na.? 

13.  It*  a  man  walk  22  M.  6  fur. 
29  rds  in  one  day,  how  far  can  he 
walk  in  6  days? 

15.  A  man  has  4  farms,  each 
containing  54  A.  3R.  23  sq.  rds.; 
how  much  in  all? 

17.  There  are  6  piles  of  wood 
each  containing  12  G.  110  s.  ft. 
1129  s.  in.;  how  many  cords  in 
all? 

19.  A  silversmith  made  12 
spoons,  each  weighing  2  oz.  3 
pwts.  8  grs.  Required,  the 
weight  of  the  whole? 

21.  A  farmer  sold  11  loads  of 
hay,  each  weighing  18  cwt.  3  qrs. 
21  lbs.  Required,  the  weight  of 
the  whole? 

23.  A  merchant  bought  10 
sacks  of  coffee,  each  weighing 
156  lbs.  12  oz.  14  drs.  Required, 
the  weight  of  the  whole? 

25.  What  is  the  weight  of  9 
packages  of  medicine, each  weigh- 
ing 3  ft).  93.  5  3.  1  9.  16  grs.? 

27.  If  it  takes  a  man  1  hr.  10 
min.  20  sec.  to  walk  a  league, 
how  long  will  it  take  him  to  walk 
7  leas^ues? 

2P.  The  earth  moves  daily  in 
her  orbit,about  59'  8'';  how  much 
does  it  move  in  9  days? 

3 1  ♦  There  are  1 8  sacks  of  flax- 
seed, each  containing  9  bu.  3  pks. 
6  qts.;  how  much  in  all? 

Remark  5.  When  the  multiplier  is 
a  composite  number  we  may  first  mul- 


tain  33  galls.  1  pt.,  how  much 
does  each  contain? 

12.  If  it  takes  24  yds.,  3  qrs. 
3  na»  to  make  7  coats,  how  much 
does  it  take  to  make  one? 

14.  If  a  man  walks  137  M. 
14  rds.  in  6  days,  how  far  does 
he  walk  in  a  day? 

16.  If  4  equal  farms  contain 
219  A.  2  R.  12  sq.  rds.,  how 
many  acres  does  each  contain? 

18*  If  6  equal  piles  of  wood 
contain  77  C.  23  s.  ft.,  1590  s. 
in.,  how  much  is  there  in  each 
pile? 

20.  If  1 2  spoons  weigh  2  lbs* 
2  oz.,  what  is  the  weight  of  each, 
each  weighing  the  same? 

22.  If  11  equal  loids  of  hay 
weigh  10  T.  8  cwt.  2  qrs.  6  lbs., 
what  is  the  weight  of  each  load? 

24.  If  10  sacks  of  coffee 
weigh  15  cwt.  2  qrs.,  18  lbs.  12 
drs.,  what  is  the  weight  of  each 
sack,  all  being  equal? 

26.  If  9  equal  packages  of 
medicine  weighs  34  lb.  3  3.  23. 
19.4  grs.,  what  is  th3  weight 
of  each? 

28.  If  it  takes  a  man  8  hrs. 
1 2  min.  20  sec.  to  walk  7  leagues, 
how  long  will  it  take  him  to  walk 
one  league? 

30.  If  the  earth  moves  in  her 
orbit  8°  52'  12"  in  9  days,  how 
far  does  she  move  per  day? 

32.  If  18  sacks  contain  178 
bu.  3  pks.  4  qts.  of  flaxseed, 
how  much  does  each  contain,  all 
being  of  the  same  size? 

Remark  6.  When  the  divisor  is  a 
composite  number,  we  may  first  divide 


When  the  multiplier  or  divisor  is  a  composite  nu.iiber,  how  may  we  proceied  7 


100 


COMMON    ARITHMETIC. 


Sect.  IX 


tiply  by  one  factor  and  then   by  the 
other.    (Sect.  IV,  Art.  4,  Obs.  4.) 

Operation  of  the  last  example. 
bu.  pks.  qts. 
9  ._  3  _  6     18  =  6  X3 
6 


first  by   one  factor,  and  then   by  the 
otner.     (Sect.  V,  Art.  4,  Obs.  3.) 

Operation  of  the  last  examj^le, 
bu.  pks.  qts.        18  =  3  X6 

3)178. „  3._  4 

[of  1 8,  or  6  sks. 

6)   59, _  2..  4  =  quantity  in  i- 


59 


2  .. 


4  =  quantity  in 
3  [6  sacks. 


Ans.  178  _.  3._  4=  quantity  in 

[6  X  3  =  18  sacks. 

33»  In  36  casks  of  wine    each 

containing  28  galls*  3  qts.  1    pt, 

3  gi.;  how  many  gallons'? 

35.  If  it  takes  2  yds.  1  qr.  3 
.na.  of  cloth  to  make  a  pair  of 
pantaloons,  how  much  will  it 
take  to  make  24  pair? 

37.  A  man  took  49  loads  of 
wood  to  the  city,  each  containing 
1  C.  15  s.  ft.,  124  s.  in.;  how 
many  cords  in  all? 

39*  A  merchant  bought  63 
kegs  of  butter,  each  weighing  32 
lbs.  14  oz.  12  drs.  Required, 
the  weight  of  the  whole? 

41.  If  a  man  walk  25  M.  6 
fur.  37  rds*  in  a  day,  how  far  can 
he  walk  in  127  days? 


Rkmark  7.  When  the  multiplier 
exceeds  12,  and  is  not  a  composite 
number,  we  multiply  each  denomina- 
tion, and  reduce  it  on  a  part  of  the 
slate  separate  from  the  multiplier  and 
multiplicand,  and  merely  set  down  the 
result  as  when  the  multiplier  is  less 
than  12. 


Ans.     9 . .  3- -  6  =  quantity  in  \ 
[of6,orl  sk. 


34.  If  36  equal  casks  of  wine 
contain  16  hhds.  34  galls.  3  qts. 
1  pt. ;  how  much  does  each  con- 
tain? 

36.  If  it  takes  58^  yds.  of  cloth 
to  make  24  pair  of  pantaloons, 
how  much  does  it  take  to  make 
one  pair] 

38.  If  49  equal  loads  of 
wood  contain  54  C.  98  s.  feet. 
892  s.  in.,  how  much  is  that  per 
load? 

40  If  63  kegs  of  butter  weigh 
1  T.  2  qrs.  24  lbs.  1  oz.  4  drs., 
what  is  the  weight  of  each,  all 
weio-hinoj  al  ke? 

42.  If  a  man  travel  3284  M. 
7  fur.  19  rds.  in  127  days,  travel- 
ino'  the  same  distance  each  day 
how  far  does  he  travel  per  day'^ 

Remark    8.      When   the  divisor  is 
greater  than  12,  and  is  not  a  compo.sil 
number,  perform  the  operation  by  long 
division. 


w 


-When  the  multiplier  exceeds  12,  anH  is  not  a  composite  number,  how  do  we 
proceed?  By  wh^it  other  method?  What  is  ihe  third  nwUxoil  Wheu  the  di- 
visor exccetls  i2,  and  is  wot  a  composite  auinberj  how  do  we  proceed? 


Art.    6.    MULTIPLICATION  AND  DIVISION  OF  COMPOUND  no's.  191 


Operation  of  the  last  example. 

M.     fur.     rds. 
25  . .  6  . .  37 
127 


Ans.    3284  ..  7  ..  19 

rds. 
37X127  =  4699 

4jU)a.^j9 


fur.  117-19  rds.  rem. 
6X127  =  762  fur,  add. 


8)879  fur. 


M.  109-7  fur.  rem. 
25X127  =  3175  M.  add. 


3284  M. 


Or,  if  the  student  prefers,  he 
may  first  multiply  by  10,  and  the 
product  thence  arising  by  10, 
which  will  give  the  product  by 
100:  then  multiply  the  product  by 
100,  by  the  number  of  hundreds, 
the  product  by  10,  by  the  num- 
ber of  tens;  and  the  first  multi- 
plicand by  the  number  of  units ; 
and  add  these  last  three  results 
to^^etlicr.     Thus — 


Operation  of  the  last  example. 

M.     fur.     rds.  [Ans. 

127)3284  ..  7  ..  19(25  ..  637 
254 

744 
635 

109  M.  remainder. 
8 

872  furlongs. 
7 fur.  aJd. 

127)879(6  fur. 

762 

1 1 7  fur.  remainder. 
40 


4680  rods. 
19  rds.  add. 


127)4699(37  rds. 
381 

889 
889 

000 

It  will  be  perceived  that  the 
only  diflference  betwe  n  this  op- 
eration, and  that  of  Ex.  2,  is 
that  in  this  operation  the  entire 
work  is  set  down,  whilst  in  Ex. 
2  it  is  not. 


192 


COMMON    ARITHMETIC, 

Sect. 

IX 

Operation  2. 

m 

^TUm^ 

127=  100 +  ! 

20  -f-  7,  or  1  hund., 

2  tens, 

7  units. 

M. 

25   . 

fur. 
.    6 

rds.              M. 
..    37  X  7  =  181    , 
10 

fur. 
..   0   . 

,.    2    . 

rds. 
.    19 

.    20 

258    . 

.    5 

..    10  X  2  =  517   . 
10 

C  2586    . .    4   . .    20  =  product  by  100. 
Add^    517   ..    2  __   20  =  pro  luct  by  2  tens,  or  20. 
{    181    ..   0   ..    19  =  product  by  7  units,  or  7. 

Ans.  3284   .,    7   ..    19  =product  by  100  +  20  + 7  =  127. 

Operation  3. 

25  M.  6  fur.  37  rds.  =  8277  rds. 
127 


57939 
16554 

8277 

1051179  rds. 
1051 179  rds.  =  3284  M.  7  fur.  19  rds.     Ans. 

Obs.  1 .  It  may  be  asked  by  some  pupils,  why  we  cannot  multi- 
ply compound  numbers,  when  the  multiplier  exceeds  12,  in  the 
same  manner  as  simple  numbers  ;  that  is,  by  multiplying  by  each 
figure  of  the  multiplier  separately,  and  writing  the  first  figure  of 
each  product  under  the  figure  by  which  we  multiply?  We  answer, 
that  we  can,  if  the  multiplicand  is  first  reduced  to  but  one  denomi- 
nation, as  in  Operation  3.  But  when  the  multiplicand  is  not  thus 
reduced,  each  denomination  must  he  multiplied  separately  hy  the  entire 
multiplier,  on  account  of  the  inequality  of  their  increase.  Did  all  the 
denominations  increase  by  the  same  ratio,  we  could  then  multiply  in 
compound  as  in  simple  numbers. 

What  is  the  difference  between  the  operations  of  Ex.  41  and  42?  When  can 
we  multiply  compound  numbers  by  multipliers  exceeding  12,  in  the  same  man- 
ner Rs  in  simple  numbers?  Why  cannot  we  proceed  iu  this  way  when  the 
multiplicand  is  not  thus  reduced?  What  would  be  necessary,  in  order  to  have 
the  same  rule  apply  to  compound  and  simple  numbers? 


Art.  6.    MULTIPLICATION    AND   DIVISION    OP    COMPOUND   No's.       193 


43.  How  much  wine  in  29 
casks,  each  containing  47  galls. 
3  qts.  2  pts.  3  gi.? 

45.  How  much  wheat  can  be 
put  in  23  casks,  each  containing 
4bu.  3pks.  6  qts.? 

47.  If  it  takes  4  yds.  3  qrs.  3 
na.  of  cloth  to  make  a  suit  of 
clothes,  how  much  will  it  take  to 
make  349  suits? 

49.  How  much  sugar  in  53 
hhds.,  each  containing  12  cwt.  2 
qrs.  23  lbs.  ? 


44.  If  29  casks  of  wine,  each 

containing    the    same    quantity, 

hold  22  hhds.  8  galls.  2  qts.  1  pt. 

j  3  gi.,  how  much  is  that  for  each? 

46.  If  23  casks  hold  ll3bu. 
2  pks.  2  qts.  of  wheat,  how 
much  is  that  for  each,  they  all 
being  of  the  same  size? 

48.  If  it  takes  1723  yds.  3  na: 
of  cloth  to  make  349  suits  of 
clothes,  how  much  does  it  take  to 
make  one  suit? 

50.  If  53  hhds.  of  sugar  con- 
tain 33  T.  14  cwt.  2  qrs.  19  lbs., 
how  much  is  that  for  each  ? 


To  find  the  difference  in  the  time  of  two  places  : 

Obs.  2.  The  earth  in  revolving  on  its  axis,  performs  one  entire 
revolution  in  24  hours,  and  as  the  circumference  of  the  earth  is  360°, 
she  makes  360°  -^-  24  =  15°  of  motion  in  1  hour  of  time,  and  60-f. 
15=4,  i.e.  1°  of  motion  in  4  minutes  of  time.  Therefore,  therp  is 
a  d'ference  of  1  hour  in  the  time  between  two  places  for  every  Xb'qf 
longitude  between  them,  and  of  4  minutes  for  every  degree.     Hence— 

To  find  the  difference  in  the  time  of  two  places : 

Obs.  3.  Multiply  the  difference  in  the  longitude  of  the  two  places ^ 
{in  degrees  and  minutes,)  by  4.-  the  result  will  he  ike  differeywe  in  tirm^ 
in  minutes  and  seconds. 

Remark.— The  student  will  bear  in  mi ud,  that  as  the  earth  revolves  from 
west  to  east,  placea  easterly  have  the  time  earlier  than  places  westerly,*    Henct— > 

Obr.  4.  To  find  the  time  of  places  easterly  add  the  difference  in 
the  time  of  the  two  places  to  the  given  time,  and  to -find  the  time  of  places 
westerly  subtract  the  difference  in  the  time  of  the  two  places,  from  the 
given  time. 

61.  The  difference  in  the  longitude  between  Boston  and  Wftst- 


How  often  does  the  earth  perform  one  revolution  on  its  axis?  How  many 
degrees  of  motion  does  it  make  in  one  hour  of  time?  Why  is  this  the  case? 
hSw  long  does  it  take  it  to  make  1°  of  motion?  What  is  the  inference  de- 
duced from  these  consideratious?  How  then  do  we  find  the  difference  in  the 
time  of  two  places?  Which  has  the  time  the  earliest,  places  easterly,  or  pUcei 
westerly?     Why  so? 

~^h?  teacher  can  best  espial*  this  by  means  of  aa  apple  or  other  rottud  8ttbst«M#i 
10 


194  COMMON     ARITHMETIC.  Scct.    IX 

ington  City  is  6°  40';  now  when  it  is  4  o'clock,  P.  M.  at  Washington, 
what  is  the  time  at  Boston? 

Operation. 

We  find  the  difference  in  the  time  of  the  two  6°       40' 

places  to  be  26  min.  40  sec;  and  Boston  being  4 

east  of  Washington,   the  time  is  earlier,    and 

consequently  it  must  be  26  min.  40  sec.  past  4         2omin.  40  sec. 
o'clock  at  Boston,  when  it  is  4  o'clock  at  Washington. 

62.  Washington,  D.  C,  being  77°  1'  30",  and  Columbus.  0., 
83°  3'  W.  longitude  from  Greenwich,  I  wish  to  know  what  time  it 
is  at  Columbus,  when  it  is  9,  A.  M.,  at  Washington? 

Ans.  35  min.  54  sec.  past  8  o'clock,  A.  M. 

63.  Beading,  Eng.,  is  about  1°,  and  Buffalo,  N.  Y.,  is  about  78° 
65'  W.  longitude  from  Greenwich.  Required — the  difference  in 
the  time  of  the  two  places. 

Ans.  5  hrs.   11   min.   40   sec. 

64.  If  it  is  9  o'clock,  30  min.  P.  M.  at  Buffalo,  what  is  the  time 
at  Reading? 

Ans.  41  min.  40  sec,  past  2  o'clock,  A.  M.  (on  the  next  morning.) 

65.  If  Buffalo  is  78°  55'  W.  longitude,  and  Tunis,  in  Africa 
about  10°  E.  longitude  from  Greenwich,  when  it  is  5  min.  20  sec. 
past  6  o'clock,  P.  M.  at  Buffalo;  what  is  the  time  at  Tunis? 

Ans.  1  min.  past  12  o'clock,  midnight. 

Remark  1.  When  one  place  is  East,  and  the  other  place  West  longitude, 
we  must  add  their  longitude  together  to  find  the  distance  between  them. 

66.  In  the  last  example,  when  it  is  26  min.  10  sec,  past  1  o'clock 
A.  M.  at  Tunis,  what  is  the  time  at  Buffalo? 

Ans.  30  min.  30  sec.  past  7  o'clock  P.  M.  {Le.  the  day  before.) 

Remark  2.  The  learner  will  observe  in  these  examples,  that  we  count  our 
time  by  half  days  or  12  hours;  and  therefore,  when  we  obtain  more  than  12 
hours,  it  is  some  hour  less  than  12,  of  the  preceding  or  succeeding  A.  M.  or  P. 
M.  part  of  the  day.  CouBequently,  if  our  given  time  is  less  than  the  diflereuce 
of  time  between  the  two  places,  and  we  wish  to  subtract  the  difference  of  time, 
we  must  add  12  hours  to  our  given  time,  and  then  subtract  as  usual. 

67.  If  a  certain  place  is  160°  E.  longitude,  and  another  place  is 
120°  W.  longitude,  and  it  is  5  o'clock,P.  M.,atthe  latter  place, what 
time  is  it  at  the  former? 

Ans.  40  min.  past  11  o'clock,  A.  M.  (i.  e.  before.) 

How  do  we  find  the  time  of  places  easterly?  Of  places  westerly?  When 
one  place  is  east,  and  another  place  is  west  longitude,  how  do  we  find  the  dis- 
tance between  them?  In  these  examples,  how  do  we  count  our  time?  What 
then  is  the  result  when  we  have  more  than  12  hours?  If  our  given  time  is 
less  than  the  difference  between  the  time  of  the  two  places,  and  we  wish  to 
•ubtmet,  how  do  we  proceed? 


■    iT-      EXCHANGE.  195 

58.  In  the  Icist  example,  when  it  is  10  o'clock,  A.  M.  at  the  for- 
mer place,  what  time  is  it  at  the  latter? 

Aus.  20  min.  past  3  o'clock,  P.  M.  (afterwards.) 

59.  What  is  tlic  difference  of  time  bi'tween  the  places  last  men- 
tioned? Ans.  5  his.  20  min. 

60.  Suppose  a  meteor  to  appear  so  high  in  the  heavens  as  to  be 
visible  at  the  same  moment,  at  Boston  71°  4'  9",  at  tlie  city  of 
Washington  77°  1'  30,"  and  at  the  Sandwich  Islands  155^*  W. 
longitude,  and  that  its  appearance  at  Washington,  be  at  45  min.  45 
sec.  past  1 1  o'clock  in  the  evening,  what  would  be  the  time  of  its 
appearance  at  Boston,  and  at  the  Sandwich  Islands. 

C  At  Boston,  9  min.  34  sec.  past  12  o'clock,  midnight  (after.) 
Ans.  <  At  the  Sandwich  Islands,  .'^3  min.  51  sec.  past  6  o'clock, 
(      P.  M.  (be'ore.) 


SECTION  X. 


EXCHANGE. 


Obs.  1.  The  method  of  finding  the  value  of  the  currency  of  one 
country  in  that  of  another,  is  called  Exchange. 

Ob-s.  2.  Currency  signifies  the  circulating  medium  of  trade, 
and  is  called  Money. 

Obs.  '?.  Money  is  said  to  have  two  values  :  an  intrinsic,  and  a 
commercial  value. 

Obs.  4,  The  intrinsic  value,  is  the  value  of  the  money  of  one 
country,  as  compared  Avith  that  of  another,  with  respect  both  to 
weigld,  and  the  purity  of  the  metal  of  which  it  is  made. 

Obs,  5.  The  commercialvcdue  is  the  compative  value  of  the  mon- 
ey of  different  countries,  with  respect  to  their  weight,  fineness,  and 
market  price. 

Obs.  6.  The  relative  value  of  foreign  coins,  (th.at  is,  the  value 
they  bear  to  each  other,)  is  determined  by  the  laws  of  the  country. 

The  value  of  a  few  of  the  most   common    foreif>n   coins   in   the 


What  is  Exchanfje?  What  d>;e3  currency  signify?  How  many  values  has 
money?  What  is  the  intrinsic  value?  The  commercial  value?  ilow  is  tho 
relative  value  of  foreign  coins  determined?  '^;    * 


196-  COMMON    ARITHMETIC.  Scct.    X 

Ukited  States  as  established  by  Act  of  Congress,  1842,  is  sbown 
in  the  following 

TABLE  : 

The  value  of  a  Pound  Sterling,  or  Sovereign,  of  England,  is  84. 846. 

The  **  a  Guinea,  **  is     5.075. 

The  **  a  Franc.  of  France  is     0.185. 

The  **  a  Five  Franc  piece,  **  is     0.93. 

The  "  a  Dollar,  of  ]^.itxico,  Peru,  and  Chili,  is     1.00. 

The  "  a  Doubloon,  of  Spain,  Mexico,  &c.,  is  15.535. 

The  "  a  Ducat  of  Russia,  is     2.297. 

Case  1 .     To  change  Foreign  Coins  to  Federal  Money. 

Ex.   1»     What  is  the  value  of  20   Pounds  Sterling,  in  Federal 
Money? 

Operation. 
One  Pound  Sterling  contains  f  4. 846,  therefore  20  ^4. 846 

Pounds  Sterling  contains  20  times  ^4.846.  20 

Ans.  $96,920 
Remark. — This  character  (£.)  nsually  stands  for  the  Pound  or  Sovereign. 

The  other  coins  are  reduced  to  Federal  Money  in  the  same  man- 
ner.    Hence — 

To  change  foreign  coins  to  Federal  Money: 

Obs.  7.     Multiply  the  given  number  of  coins,  by  the  number  of  dol- 
lars in  one  coin  as  taken  from  ike  table. 

2.  Reduce  50  Sovereigns  to  Federal  Money.        Ans.  824^,.. 30. 

3.  Change  75  Guineas  to  Federal  Money.  Ans.  $380.62^. 

4.  Change  90  Guineas  to  Federal  Money.  Ans.  $456 .  75. 

5.  Change  55  Francs  to  Federal  Money*  Ans.  '$10. 175. 

6.  Change  180  Francs  to  Federal  Money.  Ans,  ^33.30. 

7.  In  84  Five  Franc  pieces,  how  many  dollars?     Ans.  s?78. 12. 
C.  What  is  the  value  of  125  Five  Franc  pieces  in  Federal  Mon- 
ey? Ans.  ^116.25. 

9.  Change  35  Doubloons  to  Federal  Money.        Ans.  $543,725. 

10.  Change  18 Doubloons  to  Federal  Money.      Ans.  8279.63. 

1 1 .  Change  38  Ducats  to  Federal  Money.  Ans.  ^87 .  286. 

12.  Change  22  Ducats  to  Federal  Money.  Ans,  $50,534. 


Give  tho  value  of  some  of  ll*e  foreign  coins  as  shown  in  the  table?     How 
do  we  change  foreign  coins  to  Fedentl  Money? 


EXCHANGE.  197 

Case  2.     To  charige  Federal  Money  to  other  currencies . 

Remark — This  case  is  the  opposite  of  the  preceding  one,  aud  each  proves  tho 
ether. 

Ex.   L     Change  $72.69  to  Pounds  Sterling.  Ans,  15. 

Operation. 
As  1  Pound  contains  84.846.  there  will     4.848)72.690^15  Ads. 
evidently  be  as  many  Founds  in  #72.69,  as  4846 

^4,846  is  contained  in  $72.69.  

24230 
24230 


00000 


Federal  Money  is  changed  to  other  currencies  in  the  same  manner. 
Hence — 

To  change  Federal  Money  to  otlier  currencies; 

Obs.  8.  Divide  the  given  sum  by  the  number  of  dollars  it  takes 
to  make  one  of  the  coin  to  which  it  is  to  be'!;reduced,  and  point  off 
as  in  Division  of  Decimals.     (Sect.  VIII,  Art.  12.     Rule.) 

Remark.— 4  farthings  make  1  penny;  12  pence  make  1  shilling;  and29  shil- 
lings make  1  Pound  or  Sovereign,  Hence— if  thtreisa  remainder  in  reducing 
Federal  Money  to ^Pouads  St-?rling,  we  can  reduce  it  lower  as  in'Divi>ion  of 
Compound  Numbers.  (Sect.  IX,  Art.  6.  Rule  for  the  division  of  Compound 
Numbers).  Farthings  are  marked,  far.;  pence,  d.  and  shillings,  s.;  1,  2,  and  3 
farthings  are  generally  written  as  \,  i,  and  |  of  a  penny. 

2.  Change  8121.15  to  Sovereigns.  Ans.  25, 

3.  Change  $253.75  to  Guineas.  Ans.  50. 

4.  In  ^69.75  how  many  Five  Franc  pieces?  Ans.   75. 

5.  In  $217.49  how  many  Doubloons?  Ans.   14. 

6.  In  $68.91  how  mnny  Ducats?  Ans.  30. 

7.  Change  $32,83165  to  pounds,' shillings,- <'i"d  pence. 

Ans.  £6,  15s;  r)d. 

8.  Change  $21,807  to  Sterling  Money.  Ans.  £4,  10s. 

9.  Change  $52.3368  to  Sterling  Money.  Ans.  £10,  16s. 

10.  Change  $27.5*:fl  to  Sterling  Money. 

Ans.  £5,  13s.  5d.  3  far.+ 


How  do  we  change  Federal  Money  to  other  curreiici^s?  How  is  the  Pouad 
divided.?  If  there  is  a  remainder iu  reducing  Fed^jral  Mone}'  lo  Pouuda;  Stef- 
Ijng^  how  may  we  proceed? 


198  COMMON    ARITHMETIC.  Scct.  XI 

SECTION  XI. 

'   •  ANALYSIS. 

Article   1.     Definitions  and  Mental  Exercises. 

Obs,  1.  Aiialy sis  sigrtijies  the  separation  of  a  body  into  parts,  and  is 
used  in  mathematics  for  the  developement  and  illustration  of  principles, 
and  may  also  be  apjjhed  to  the  solution  of  questions  voilh  great  practical 
advantage.  In  the  preceding  sections  we  have  analyzed  Principles; 
in  this  section  we  design  to  analyze  practical  questions,  or  Pro- 
blems. 

NoTK. — We  often  liear  peoplft  tpil  of  "doing  sums  in  their  head."  They  in 
ill  reality,  solve  their  quesliouH  Ly  Analysis,  or  the  common  sense  rule. 

Obs.  2.  This  method  is  of  extensive  application,  and  great  util- 
ity. It  strengthens  the  mental  faculties,  and  accustoms  the  pupil  to 
habits  of  close  thinking,  and  with  a  little  practice  he  can  become  very 
expert  at  it. 

No  particular  diiection  can  be  given  for  solving  Q.\i^rj  question  in 
analysis.  The  following  may  perhaps  answer  for  general  directions, 
but  the  learner  must  try  and  think  for  himself  and  not  depend  too 
much  on  others. 

Obs.  3.  Examine  the  question  attentively;  reason  according  to  the 
nature  of  the  sum,  and  work  as  common  sense  directs. 

Remark. — The  operation  of  solving  questions  by  analysis  is  called  an  Ana- 
lytic Solution. 

Obs.  4.  In  soloing  questions  analytically,  'we  generally  reason  f  rem 
the  given  number  to  1,  and  then  from  1  to  the  required  numher. 

MENTAL    EXERCISES. 

1 .  If  5  tans  of  hay  cost  830,  what  will  8  tons  cost? 

Ans.  48. 
Solution.— li  5  tons  cost  $30,    1    ton   will  cost  J  of  $30,  or  ^1?,- 
and  8  tons  will  cost  8  times  $6  or  $48. 

2.  If  3  firkins  of  butter  eost  $15,  what  cost  4  firkins?  6  ;  8;  iO; 
12? 


What  does  an  alysi.«!  signify  ?  To  what  is  it  used  in  malheniatics?  To  what 
may  it  be  applied?  How  do  people  who  "do  sums  in  their  head"  in  realrfy 
solve  their  questions?  What  advantages  are  derived  from  this  method?  Wliut 
directions  are  given  for  analyzing?  What  is  the  operation  of  solving  a  ques- 
tion by  analysis  called?  How  do  we  generally  reason  in  solving  que«.tion3 
analytically? 


Art.  1.  ANALYSIS.  199 

3.  If  a  man  can  earn  $14  in  7  days,  how  much  can  he  earn  in  3 
days?  4;  5;  6;  8;   10;   12;  15? 

4.  If  a  man  spend  $18  in  6  days,  how  much  does  he  spend  in  4 
days?     3;  5;  9;   10;  12? 

5.  If  a  b  )y  walk  72  miles  in  6  days,  how  far  does  he  walk  in  6 
days?     8;  10;   12? 

6.  If  4  men  can  do  a  piece  of  work  in  3  days,  how  long  will  it 
take  6  men  to  do  the  same  work?  Ans.  2  days. 

Solution. — If  it  takes  4  men  3  days  to  do  the  work,  it  will  take  1 
man  3X4=12  days  to  doit.  Again  if  it  takes  1  man  12  days  to 
do  it,  6  men  can  do  it  in  J  of  the  time,  or  2  days. 

7.  If  it  takes  3  men  8  days  to  do  a  piece  of  work,  how  long  will 
it  take  2  men  to  do  it?     4;  5;  6;  8;   12;  24? 

8.  If  4  horses  consume  a  stack  of  hay  in  12  weeks,  how  long 
would  it  keep  2  horses?     6;  8;   12;  24? 

9.  If  a  certain  pasture  keeps  6  cows  6  weeks,  how  long  will  it 
keep  4  cows?     8;  9;   12;   18;  36; 

IJ.  Two  boys  were  counting  their  money:  one  said,  "I  have.  50 
cents;"  the  other  said,  **I  have  j  as  much  as  you."  How  much 
had  the  latter  boy? 

Soliition. — J  of  50  cents  is  10  cents,  and  J  is  4  times  as  much  as 
J,  and  4  times  10  are  40.  Ans.  40  cents 

11.  A.  and  B.  were  talking  of  their  ages:  A.'s  age  was  63  years, 
and  B,'s  age  was  y  of  A.'s  age.     What  was  the  age  of  B.? 

12.  John  and  Thomas  were  playing  marbles:  John  had  24,  and 
Thomas  had  |  as  many  as  John,     How  many  had  Thomas? 

13.  A.  has  96  sheep;  B.  has  —  as  many  as  A.  How  many 
hasB, 

14.  One  man  has  132  hogs,  another  man  has  ^^  as  many;  how 
many  has  the  latter  man? 

15.  Tho  lias  has  120  bushel  of  oats;  jq-  of  Thomas'  is  equal  to  y 
of  James';  how  many  bushels  has  James?  Ans,  126. 

Solution. — yV  of  120  is  12,  and  y^  of  120  is  9  times  12  or  108; 
then  108  is  ^  of  James';  if  108  is  7,  y  is  i-  of  108  or  18,  and  ^  or 
the  whole  quantity  7  times  18  or  126. 

16.  Two  boys  were  talking  of  their  money;  one  says,  "I  have  12 
cents;"  the  other  says,  "J  of  your  money  is  equal  to  |  of  mine;" 
how  much  had  he? 

17.  Two  boys  talking  of  their  ages,  one  said  "I  am  15  years 
old;"  the  other  said,  J  of  your  age  is  equal  to  y  of  mine".  How 
old  was  he? 

18  A.  has  14  sheep;  4  of  his  number  is  equal  to  |  of  B.'s;  how 
many  has  B.? 


200  COMMON    ARITHMETIC,  ScCt.    XI 

19.  Charles  has  $36;  ^  of  his  is  equal  to  —  of  Francis';  how 
much  has  Francis? 

20.  A  man  bought  a  horse  and  paid  $30  down,  which  was  I  of 
the  price  of  him.     What  was  the  price  of  the  horse? 

Ans.  $54. 
Solution. — If  30  is  |-  of  a  number,  \  will  be  \  of  30  or  6,  and  J, 
or  the  number  itself  will  be  9  times  6,  or  54. 

21.  A  note  has  run  45  days,  which  is  I  of  the  time  it  has  to  run; 
how  long  has  it  to  run? 

22.  A  man  bought  a  waggon,  paying  $36  down,  which  was  |-  of 
the  price  of  it.     Required,  its  price? 

23.  James  bought  some  iron,  paying  $48  [^trade;  this  was  J  of 
the  cost  of  it;  what  did  it  cost? 

24.  A  lady  purchasing  some  goods,  paid  $72  down;  this  was  /-g- 
of  her  purchases;  how  much  did  she  trade? 

25-  A  man  in  trading,  paid  down  $84;  this  was  -j  of  his  bill;  re^ 
quired  the  amount  of  his  bill,  and  how  much  he  had  yet  to  pay? 

26.  Frank  and  Henry  were  speaking  of  their  money.  Henry 
says  "I  have  84  cents;"  says  Frank  "I  have  y  as  much  as  ycu,  and 
am  going  to  buy  lemons  with  it,  at  4  cents  apiece;"  how  many  can 
he  buy?  Ans.   18. 

Solution. — I  of  84  is  72,  the  number  of  cents  Frank  had.  Then 
as  the  lemons  are  4  cents  apiece,  he  can  buy  72 -f- 4---  18  lemons 
for  72  cents, 

27.  A  man  has  y 2  of  $144;  how  many  yards  of  cloth  can  he 
buy  at  $6  per  yard? 

28.  Henry  has  yj  of  $132;  how  many  steers  can  he  buy  with  it 
at  $9  apiece  ? 

29.  A  nan  having  fV  of  $120,  bought  9  coats  with  it;  how  much 
were  they  apiece? 

30.  Chailes  h:;d  a  note  for  $96;  he  obtained  -  of  it,  and  spent  it 
for  sheep  at  $3  apiece;  how  many  did  he  buy? 

31.  WilHam  spent  $14  for  books,  which  ^yas  y  of  all  he  had  ;  he 
spent  the  remainder  for  7  yards  of  cloth;  what  did  the  cloth  cost  him 
per  yard?  Ans,  $5. 

Solution. — If  $14  was  f  of  all  he  had,  \  would  be  \  of  $14,  or 
$7,  aud  I  would  be  7  times  $7,  or  $49.  Then  as  he  spent  $14, 
he  would  have  $49-$14=$35  remaining;  with  this  he  bought  7 
yards  of  cloth,  which  must  have  cost  him  $35-7-7= $5  a  yard. 

32.  A  farmer  paid  $30  for  goods,  which  was  |  of  his  bill;  he 
paid  the  balance  in  wheat  at  %\\  per  bushel;  how  many  bushels 
did  it  take? 

33.  A  man  spent  $150  for  sheep,  which  was  /,-  of  all  he  had: 
he  spent  the  remainder  for  land  at  $10  ;  n  acre;  how  many  acres 
did  he  buy? 


Art.   ^<.  ANALYSIS.  301 

34.  A  man  bought  a  iiorse  paying  ^36  down,  ■wliicli  was  J  of 
what  he  was  to  .pay;  he  paid  the  remainder  in  flour  at  <$5  a  barrel^ 
how  many  barrels  did  it  take? 

35.  Lewis  spent  ^72  for  cloth,  which  was  yy  of  all  he  had,-  he 
«pent  the  remainder  for  books  at  $4  apiece:  how  many  books  did 
he  buy? 

36.  Silas  spent  24  cents  for  a  slate,  which  was  |  of  all  he  had; 
he  bought  10  lead  pencils  with  what  was  left;  how  much  were  the 
pencils  apiece? 

37.  I  of  24  is  how  many  times  10?  Ans.  2. 
Solution.—^  of  24  is  20;  20-r-  10  =  2;  hence,  20  is  twice  10. 

^38.  J  of  48  is  how  many  tfenes  6? 

39.  1^  <5f  54  is  how  many  times  3? 

40.  J  of  56  is  how  mmiy  tirfees  4? 

41.  48  isf"of  how  many  times  6?  An's.  "9. 
Solution. — ^We  first  inquire,  48  is  ~  of  what  number?     K  48    is 

~,  one  ninth  is  j  of  48,  or  6;  and  J  or  the  number  itself  is  9  times 
6,  or  54.  Then  54  is  how  many  times  6?  51  -r-6  =  9,  that  is  54 
is  9  times  6? 

42.  63  is  yV  of  hcrwmany  times  15? 

43.  1 08  is  /o  of  how  many  times  30? 

44.  6  times  6  is  J  of  what  number?  Ans.  81. 

45.  8  times  9  is  |  of  how  many  times  3  times  4?  Ans.  7. 

46.  3  times  30  is  y  „  of  how  many  times  4  times  5?      Ans,  5. 

47.  J  of  36  is  y\  of  what  number?  Ans.   100. 


J 


Solution.---^  of  36  is  30;  30  is  /„  of  100. 

48.  "  of  42  is  4  of  what  number? 

49.  I  of  48  is  i  of  how  many  times  3?  Ans,   U. 

50.  4  of  72  is  /y  of  how  many  times  5  times  5? 

Ans,  2. 

Article  §.     Exercises  for  tHE  Slats. 

1.  If  25  yds.  of  cloth  cost  8100,  what  will  40  yds  cost? 

Ans.  $160. 
Solaiion.—lf  25  yds.  cost  $100,  1  yd.  will  cost  $100-^-25  =$4. 
Then  40  yds.  will  cost  40  times  as  much  as  I  yd..,  and  84  X  40  = 
$160,  Ans, 

Remark. — la  solving  such  qnesdons,  the  learner  can  often  shorten  the  ©per- 
•^tloa,  by  merely  expressing  the  work  and  cauceliug.     Thus-— 


How  can  the  operation  in  analysis  often  be  fihort^ned? 
.  10a 


202'  cowsfoN  ARi-fHMETie.  S'ect.  XI 

Wliert  tHfe  learner  performs  the  operatroir  by  can-  ^5      X^^-^^ 
celation;  (or  any  other  way)  he  should  be  re(|iri?ed  40 

to  give  his  reason  for  every  step   he  takes.     It  is  — ~ 

a  matter  of  unimportance,  haw  or  where  he  sets  bis  !$160  Ans- 

figm-esvif  he  thoroughly  u»derstands  his  subject 

Note. — -Questions  of  this  kind  are  usually  worked  bj*  ff^i^tripic-  Proportion^ 
ortheRuie  of  Three,  which  i®  expfanied  iivthe  next  section.  Business  men, 
ii9wevep,^et?eraily  solve  them  by  analysis. 

2:.  What  cost  48  cows,  if  20  cows  cost  8240?  Ans.  8576.- 

3.-  \\niat  cost  48  lbs.  of  coff^^e,  if  60  lbs,  osst  $6.60? 

Ans.  $6.28. 

4.  It  I  pay  $31.20  for  the  use  of  $520  a  certain  time,  how 
much  must  i  pay  for  the  use  of  $7*20  the  saaie  time? 

Ans.  $43.2'0. 

5.  If  a  stage  travel  90  miles  m  1-5  hours,  hmv  masy  rai^Jes  will  it- 
travel  in  36  hours?  Ans.  216. 

ei     6'.  What  cost  850  biishels  of  barley  i-f  600  bushels  cost  ^^30u? 
tf^--'^-  Ans.  $425. 

7.  If  25  horses  cat  112)^  bushels  of  oats  in  a  week,  how  many 
Ixuehels  will  14  horses  eat?  Ans.  63. 

8.  If  416  lbs.  of  wool  cost  $104,  what  wHI  599  lis-,  cost? 

Ans.  $149.7-5. 

9.  If  48  tons  of  hay  cost  $720,  what  will  36  tons  cost? 

Ans.  $540. 

10.  If  araili*oad  car  travels  r54mires  in  1 1  hours,  how  far  will 
st  travel  in  18  hours?  Ans.  252  miles. 

n.  A  man  can  do  a  piece  of  work  in  8  days,  laboring  12  hours^ 
per  day:  how  long  wiU  it  take  him  when  he  labors  but  9  hours  per 
day?  Ans.   10§  days. 

12.  If  4  lbs.  of  coffee  cost  $0.50,  what  will  40  yds.  of  calicw  cost,, 
if  7 J  lbs.  of  coffee  are  worth  5  vds.  of  calico? 

Ans,  $7.50. 

13'.  If  12  men  can  do  a  piece  of  work  in  24  days,  how  long: 
would  it  take  30  men  to  do  the  same  >orL? 

Suggestion.— li  12  men  do  the  work  in  24  days,  it  will  take  f 
Man  12  times  as  long  to  do  it,  and  30  men  will  do  it  in  ~  of  the 
^i'me  in  which  one  man  could  perform  it.  Ans.  9 J  days. 

14.  If  18  men  eat  a  barrel  of  flv'rur  in  0  days  how  long  will  it 
last  45  men?  Ans  2>~  days. 

r5.  If  a  certain  quniility  of  oats  kst  IS  horses  18  d?iys,  how 
long  will  it  last  28  horses?  Ans.    lOf  days. 


Is  it  a  matter  of  import:  r.ce  wl^rc  sr  l.sw  tile  figures  ar*  Ect  if  the  learner 
Ti.nclerstands  his  suhjecl?  '"^iwoM 


Art.  2»  AJTALYsts.  203 

16.  If  a  barrel  of  flour  lasts  8  men  30  days,  how  long  will  it  last 
18  men?  Ans.   ISj  days. 

17.  I(  16  men  can  do  a  piece  of  work  in  12  days,  how  long  will 
it  take  24  men  to  do  it?  Ans,  8  days, 

18.  If  6  stacks  of  hay  keep  40  cattle  90  days,  how  long  will  they 
keep  50  cattle?  Ans.  72  days. 

The  learner  will  perceive  that  the  number  of  stacks  is  not  to  be 
regarded  in  solving  this  question. 

19.  If  \  a  bushel  of  wheat  cost  ^0.50,  what  will  J  of  a  bushel 
cost?  "  Ans.  $0.75. 

Suggestion. — If  \  a  bushel  cost  $0.50,  a  bushel  will  cost  twice  as 
much,  and  J  of  a  bushel  will  cost  J  as  much  as  a  bushel. 

20.  If  f  of  a  yard  of  cloth  cost  $3.20,  what  costs  J  of  a  yard? 

Ans.  $3.50. 

21.  If  4  of  an  acre  of  land  costs  $18,  what  costs  |  of  an  acre? 

Ans.  14. 

22.  If  t"o  of  a  ton  of  hay  costs  $10.80,  what  will  |  of  a  ton  costi 

Ans.  10. 

23.  If  J-  of  a  ton  of  coal  costs  $4.90  what  will  y^  of  a  ton  cost? 

Ans.  $5.67. 

24.  A  merchant  bought  a  load  of  wheat,  J  of  which  cost  $24, 
and  afterwards  sold  ~-^  of  his  load  at  cost;  how  much  did  he  get 
for  what  he  sold?  Ans.  $9, 

25.  A  has  420  sheep;  \  of  A's  is  equal  to  4  of  B's;  how  many 
has  B?  ^      Ans.  294. 

26.  A  man  bought  16  yds.  of  cloth  for  $60;  if  he  should  sell 
J  of  it  at  cost,  how  much  would  he  get  for  it?  Ans.  $37.50, 

27.  If  a  man  pay  $33.50  for  26  j  days  work,  how  much  must  he 
pay  for  67 yo  days  work?  Ans.  $84. 

28.  If  a  man  ride  645  miles  in  18y  days,  how  far  can  he  ride  in 
29j  days?  '  Ans.   1036  miles. 

29.  If  7  lbs.  of  tea  cost  $5i,how  much  will  12  lbs.  cost? 

Ans.  $9. 

Suggestion,— ^6\  =  $V  •  If  7  lbs.  cost  $ V »  1  lb  will  cost  \  of 
$y ,  and  12  lbs  will  cost  12  times  as  much  as  1  lb. 

30.  If  15  acres  of  land  cost  $131^,  what  will  43  acres  cost? 

Ans.  $376.25. 

31.  If  it  cost  $2 J  to  build  112  rds,  of  fence,  how  much  will  it 
cost  to  build  216  rds.?  Ans.  $5.54^f . 

32.  If  15  pair  of  side  combs  cost  $0.93|,  what  cost  27  pair? 

Ans.  $1.68|^, 
33    If  I  of  a  bushel  of  wheat  costs  VJ  of  a  dollar,  what  costs  | 

of  a  bushel? 

Solution. — If  S  of  a  bushel  costs  y|,  J  will  cost  ^  as  much,  or  /j 

»=  i  of  a  dollar,  and  a  bushel  will  cost  6  times  as  much,   or  $1. 


204  COMMON    ARITHMETIC.  Scct.  XI 

Again,  if  a  bushel  costs^l,  |  of  a  bushel  will  cost  |  of  a  dollar, 
or  ^0.75. 

34.  If  J  of  a  yard  of  cloth  costs  y^  of  a  dollar,  what  costs  ~  of  a 
yard?  Ads.  $;0.52. 

36.  If  Yo  of  a  farm  costs  $2100,  what  costs  |-  of  the  same  farm? 

Ans.  ^2000. 

36.  If  j%  of  a  ship  costs  $48000,  what  is  ||  of  her  worth? 

Ans.  iSeCOO. 

37.  Suppose  an  army  of  1500  men  have  120000  lbs.  of  bread; 
how  long  will  it  last  them,  allowing  each  soldier  1^  lbs  per  day? 

Ans4  64  days. 

38.  Suppose  the  above  named  army  should  lose  1 20  barrels  of 
bread,  each  containing  250  lbs.;  how  much  must  be  taken  from 
each  soldier's  allowance  per  day,  in  order  that  the  remainder   mat 

last  the  same  time,  and  how  much  will  be  the  allowance  of  each 
soldier  per  day  after  the  reduction  is  made? 

Ans.  Each  soldier  loses  6oz.  per  day,  and  has  15oz.  remaining* 

39.  Suppose  an  army  of  1500  men,  having  lost  \  of  their  bread, 
were  obliged  to  subsist  upon  16  oz.  per  day  f <  r  64  days;  had  none 
of  their  bread  been  lost,  they  would  have  had  1^:  lbs.  per  day  for  the 
same  time;  how  much  bread  had  they  at  firsts  and  how  much  was 
lost? 

.        {  They  had  at  first  1 20000  lbs. 
^^®*  \  They  lost  30000  lbs. 

40-  Suppose  the  allowance  of  an  army  of  1500  men  was  short- 
ened 5oz.  per  day  for  64  days  in  consequence  of  apart  of  their 
bread  being  spoiled,  and  the  amount  spoiled  to  be  \  of  the  whole; 
what  was  the  whole  amount  both  good  and  bad,  the  amount  spoiled 
and  the  allowance  of  each  soldier  per  day,  both  before  and  after  their 
bread  was  spoiled? 

r  Total  amount  1200C0  lbs. 
j^  g  J    Amount  spoiled  30000  lbs. 

*  I    Daily  allowance  of  each  soldier  at  first  l|  lbs. 
(^  Daily         *'  "  **  after  apart  was  spoiled  15  oz^ 

Note. — Questions  similar  to  the  four  preceding,  and  the  21  following  ones, 
are  generally  solved  by  reasoning  from  one  statement  to  another,  without  ref- 
erence to  any  particular  rules. 

41.  A  man  owning  J  of  a  farm,  sold  |  of  his  share  for  $2800; 
required  the  worth  of.  the  farm?  Ans.  $4800. 

Solution. — As  he  sold  ~  of  his  share,  he  sold  |  of  -J-  of  the  entire 
farm,  f  of  1  ^  yV»  (Sect.  VIII,  Art.  3,  Obs.  10  or  1 1.)  Then 
$2800  is  ^2  of  the  worth  of  the  farm,  and  yV  must  be  \  of  $2800, 
or  $400,  and  the  entire  farm  must  be  worth  12  tiines  as  much  as 
fV  or  $400  X12=$4800.    ,  ...^  ..^,^«  ..  ^^  ,^.. 


Art.  2«  ANALYSIS. ;. .,  205 

42.  A  mercliant  owning  j  of  a  store,  sold  j  of  his  shal-e  for  ^15- 
000;  what  was  the  worth  of  the  store?  Ans.  ^25000 

43.  A  farmer  owning  |-  of  a  field  of  wheat,  sold  ^  of  his  share 
for  J$100;  what  was  the  field  worth  at  this  rate?  Ans.  ^S70. 

44.  A  cidtern  holding  400  gallons  is  to  be  filled  with  water.  By 
one  pipe  30  gallons  run  in  in  an  hour,  whilst  by  another  pipe  10  gal- 
lons run  o  't  in  an  hour;  if  both  are  left  open,  how  long  will  it  take 
the  cistern  to  fill? 

Solution.— ~K^  30  galls,  run  in  in  an  hour,  and  10  galls,  run  out 
in  the  same  time,  it  tills  30 — 10  =  20  galls,  per  hour.  Then  as  it 
holds  400  galls,  it  will  take  it  400  -H  20  =  20  hours  to  filh 

Ans.  20  hours. 

45.  A  boat  traveling  up  stream  is  driven  by  steam  at  tlie  rate  of 
1 4  miles  per  hoUr,  whilst  she  is  retarded  by  the  current  ^\  miles  per 
hour;  how  long  will  it  take  her  to  travel  418  miles? 

Ans.  44  hours. 

46.  A  vessel  sprung  a  leak  at  sea,  and  it  was  not  discovered  Until 
she  had  215  gallons  of  water  in  the  hold;  the  pumps  empty  5  gal- 
lons per  minute,  whilst  the  leak  lets  in  2^-  gallons  per  minute;  how 
long  at  this  rate  would  it  take  to  empty  the  ship  ? 

Ans.  86min.=l  hr.  26  tain. 

47.  A  man  earns  ^9  per  week,  and  spends  83.50  per  week;  how 
much  can  he  save  in  66  weeks?  Ans.  ^363. 

48.  A  cistern  can  be  filled  by  one  pipe  in  10  hours,  and  by  anoth- 
er pipe  in  1 2  hoi^j^s,.  If  both  are  left  open  how  long  will  it  take  the 
cistern  to  fill?  ..\        '  Ans.  S^y  hours. 

Solution.— 1{  a  pipe  fills  it  in  10  hours,  it  will  fill  ~  of  it  in  1 
hour.  Again,  if  a  pipe  fills  it  in  12  hours  it  will  fill  -^  of  it  in  1 
hour;  therefore  both  will  fill  y o  +  yV  ==  H  of  it  in  an  hoUr.  Then 
to  fill  the  whole  cistern,  or  |J,  it  will  take  60  ■—  11  ^  5/y 
hours. 

49.  A  cistern  has  a  pipe  which  will  fill  it  in  8  hours,  and  another 
that  will  empty  it  in  10  hours.  If  both  are  left  open,  how  long 
will  it  take  the  cistern  to  fill?  Ans.  40  hours. 

S ugjestion, -^li  fills  \  full  in  1  hour,  and  emties  ^-^  in  the  same 
time;  hence,  it  gains  j-^yV  =  yV  of  its  contents  in  an  hour. 

50.  A.  can  do  apiece  of  work  in  6  days,  whilst  it  will  take  B.  8 
days  to  do  the  same  work.  If  both  work  together,  in  -what  time 
will  they  do  it?  Ans.  3y  days. 

51.  A.  can  do  apiece  of  work  in  9  days;  B.  in  12  days,  and  C. 
in  15  days;  how  long  would  it  take  them  all  working  tosrether,  to  do 
it?    ,  Ans.  3|f 

52*'  A.  can  do  a  certain  piece  of  work  in  12  days;  B.  in  16  days; 
C.  in  20  days;  and  D.  in  24  days;  in  what  time  will  they  all,  work- 
ing together,  do  it?:  y*  lo  5  l^m  \.^3fi  Ans.  4J^|  days. 


206  COMMON  arWhmetic.  Sect.  XI 

53.  A.  anl  B.  together  can  plow  a  ccrtaini  piece  of  ground  in  10 
days.     A.  can  do  it  alone  in  16  days;  in  what  time  can  B  do  it? 

Ans.  26|  days. 

64.  A  nvMi  and  his  wife  eat  the  floar  of  a  bushel  of  wheat  in  12 
days  ;  the  woman  would  eat  it  alone  in  27  days;  in  what  time  would 
the  man  eat  it?  Ans.  21 J  days. 

55.  A  cistern  liolding  600  gallons,  has  two  pipes  which  together 
will  let  in  1 40  gallons  of  water  in  3  hours,  it  has  also  one  pipe  which 
will  'et  out  l5o  gallons  in  5  hours;  If  all  are  left  open,  how  long 
will  it  take  the  cislern  to  fill?  Ans    39  /—  hours. 

56.  A.  B.  and  C.  can  do  a  piece  of  work  in  15  days  ;  A.  and  B. 
can  do  it  in  24  days;  in  what  time  can  C.  do  it  alone? 

Ans.  40  daysv' 

57.  James  and  Henry  wish  to  divide  25  cents  between  them  so 
that  Henry  may  have  5  cents  more  than  James;  how  much  musti 
each  have? 

Solution. — If  Henry  takes  the  5  cents,  he  is  to  have  more  than 
James,  it  is  evident  that  the  remai  )der  should  be  divided  equally 
between  them.  Then  25 — 5=20;  20-r-2  =  10  cts.,  James' 
share,  ^  nd  10 -j- 5=  15  cts.;  Henry's  share.  (Sect.  VI,  Art  1, 
Gbs.  14.)     Proof,  15  +  10  =  25. 

58.  Divide  36  marbles  between  two  boys,  giv'ing  one  10  more 
than  the  other?  Aiis.   13  and  23. 

59.  A  man  left  84500  to  be  divided  between  two  children,  giving 
one  ^800  more  than  the  other;  what  was  the  share  of  each? 

Ans.  One's  share  ^1850:  the  other's  82650. 

60.  A  man  left  his  property  amounting  to  $7500  to  be  divide(^ 
between  his  wife,  son,  and  daughter,  giving  the  wife  81200  more 
than  to  the  son,  and  to  the  son  8800  more  than  to  the  daughter.—- 
What  was  the  share  of  each? 

.        i  Wife's  share,  83566.66|;  Son's  82366.66|; 
^^^-  ^Daughter's  81566  66|. 

61.  Divide  81200  between  four  person,  giving  A.  8200  more 
than  B;  B  8150  more  than  C;  and  C  8100  more  than  D. 

Ans.  A's  share  8550;  B's  8350:   C's  8200;    D's  8100. 

62.  What  cost  40  yds.  of  cloth,  at  80.25  per  yard? 

Ans.  810. 

SoltUion. — 'Had  the  cloth  been  81  per  yard,  it  would  evidently 
bave  cost  840.  Now  80.25  is  \  of  81,  therefore  it  will  cost  \  as 
many  dollars  as  there  are  yards;  and  \  of  40  is  10. 


Art.  2- 


•>^'^^-   ANALYsr 


207 


Obs,  1.  The  solution  of  questions  hy  ial:iny  i^ayts ,  as  in  this  ex* 
ample,  is  called  Practice. 

Obs.  2.  The  chief  advantage  derived  from  this  method  of  op- 
era^^ing,  is,  that  it  generally  contracts  the  operation,  thus  often  ennahling 
the  learner  to  solve  questione  mentally  which  ivoidd  otherwise  require  the 
tcse  of  the  slate. 

Obs.  3.  Any  number  that  forms  an  exact  part  of  anotlier  num- 
ber, is  called  an  aJiqiwt  jMrt  of  that  number.  Thus,  4  is  an  aliquot 
part  of  12;  50  cts.  of  §1.00;  tfec. 

Remark. —  It  will^be  perceived  that  the  tefnis  aliquot  parts,  component  purls, 
a.nd factors,  are  synonymous;  that  is,  tliey  express  the  same  meaning. 

To  obtain  the  cost  of  any  number  of  articles,  -when  the  price  of 
1  is  an  aliquot  part  of  a  dollar  : 

Obs.  4.  Divide  the  given  number  of  articles  by  that  pari  of  ^\  at 
which  a  single  article  is  priced:  the  residt  will  be  the  answer,  iti  doU 
lars. 

Remark, — If  there  is  a  remainder  after  dividingj  tlie  quotisnt  n)ay  be  ex- 
tended to  cents,  and  mills,  by  annexing  ciphers. 

Obs.   5.     Practice  is  chiefly  used  in  computing  the  iMce  of  varicui 
articles  of  trade;  as  groceries,  cloth,  land,  grain,  <fec. 
;^  The  aliquot  parts  of  8 1  are  shown  in  the  following 

TABLE. 

cents is 


20 
25 

50 


12^  cents. 

12^  -  . 

25  -  . 

25  -  . 

37i  "  . 

62i  "  . 

75  «  - 

b6|  -  . 


42S^J^ 


lof 

6 


,  .  f 
iot 


87i 


IS 


m. 

of-- - .,_.    ^1. 

\  of  i?z- it*  ^^-^  1^1  it    $  1 . 

81. 

$1. 

IS  ^or ._..-.._  ^1. 

also:  -niBd/. 

is  *;  of-. __-    ._.-  80.25. 

is  J:  of eo.50, 

is  ^of J5>0.50. 

is^of SO. 75. 

isiof {$0.75. 

is ' 

isi=i+iof 

is|  =  i+iof 9\. 

=  j  +  i  +  T.  orl-lof«l. 


|  =  i+lof «S). 

=  -■'•"«■         $1. 


What  is  practice?  What  is  the  advantage  derived  from  this  method  of  op- 
erating? What  is  an  aliquot  part  of  a  number?  Give  examples.  What  is  the 
difference  in  the  meaning  of  the  terms  aliquot  parts,  compoDfUt  parts,  and 
fiictors'i  ilow  do  we  obtain  the  cost  of  any  number  of  articles  when  the 
priceof  1  is  auMiquot  part  of  a  dollar?  If  there  is  a  remninder  how  do  we 
proceed?  For  what  is  practie*  chiefly  used?  Give  examples.  Give  •om»  of 
the  aliquot  parts  of  $l-r 


J^OS  COMMON   ARlTriMEtlC.  Sccl.    Xl 

63.  What  cost  72  lbs.  of  coffee,  at  12^  cents  a  pound? 

Ans.  $9-. 

64.  What  cost  68  yds.  of  muslin,  at  12^  cents  a  yard?- 

Ans.  ^8.60w 

65.  W^hat  cost  75  slates,  at  16|  cents  apiece? 

Ahs.  $12.50. 
rH.^6-  What  cost  llObookSj  at  20  cents  apiece?  Ans.  ^22. 

67.  What  cost  50  palm  leif  hats  at  25  cents  apiece? 

Ans.  ^12.50. 

68.  What  cost  45  bushels  of  potatoes,  at  33j  cents  per  bushelf 

Ans,  815. 

69.  What  cost  1 200  bushels  of  barley,  at  37^  cents  per  bushel? 

Ans.  $450. 

Operation. 

$1200  =  cost  at  $1. 


i:^  cost  at  25  ctSi 
ai50  =  costatl2^cts. 

$450  =  cost  at  37^  cts 


The  cost  at  $1  per  bushel,  would  be     4 
f  1 200,  the  cost  at  25  Cts.  per  bushel 
would  be  i-  of  $1200  =  $300;  the  cost     2 
at  12^  cts.  per   bushel  would   be  \  as 
much   as   at  25  ct^.,   or  \  of  $300  = 
$150;  then  the  cost  at  37^- cts.  per    Ails. 

bushel  would  be  equal  to  the  cost  at  25  cts.*-|-  the  cost  at    \9.\  cts, 
or  $300  +  $150  =  $450,  because  37^  cts.  =  25  cts.  +  \2\  cts. 

70;  What  cost  400  pair  of  socks,  at  37^  cts.  a  pair? 

Ans.  $150. 

71.  What  cost  247  yds.  of  cloth,  at  50  cts.  a  yard? 

Ans.  $123.50. 

72.  Wh^t  cost  471  bushels  of  rye,  at  62i-c:s  a  bushel? 

Ans.  $294,371^, 

73.  What  cost  600  bis.  of  cider,  at  66|  cts.  per  barrel? 

Ans.  $400. 

74.  What  cost  300  lbs.  of  tea,  at  75  cts.  per  pound? 

Ans.  $225. 
1(5.  What  cost  400  yds.  of  satinet,  at  87^  cts.  per  yard? 

Ans.  $350. 
76.  What  cost  150  acres  of  land,  at  $8*25  per  acre? 

Ans.  $1237.50. 
Suggestion.— m.'^B=z  $8^;  therefore  multiplying  150  by  Si,  it 
s  evident  we^shall  have  our^ansW-er  in  dollars.     Hence — 

When  the  pfice  of  an  article  is  in  dollars  and  cents, 

Obs.  6.     Multiply  the  number  of  articles,  hy  the  numhef  of  dollars ^ 
take  aliquot  parts  for  the  cents,  and  add  the  several  resulti  together. 

Explain  the  solution'of  Ex.  69.    When  the  price  of  an  attiele  H  in  dollard 
and  Cents  how  do  we  proceed? 


Art  2.  '^^^^  AifALYSis.  s  209 

77.  What  cost  250  bushels  of  wheat,  at  81 .  12^  per  bushel? 

Alls.  8281.25. 

78.  What  cost  360  yds.  of  cloth,  at  $2. 16|  per  yard? 

Ans.  ^780. 

79.  What  cost  480  silver  pencils,  at  $2. 25  apiece? 

Ans.  81080. 

80.  What  cost  260  gallons  of  wine,  at  82.33i-  per  gallon? 

Ans.  8606. 66|. 

81.  What  cost  80  yds  of  cloth,  at  84.37^  per  yard? 

Ans.  8350. 

82.  What  cost  160  acres  of  land,  at  812. 50  per  acre? 

Ans  82000. 

83.  What  cost  40  sacks  of  coffee,  at  812.62^  per  sack? 

Ans.  8506, 

84.  What  cost  72  bis.  of  flour,  at  87.66|  per  barrel? 

Ans.  8552. 

85.  What  cost  120  acres  of  land,  at  85.75  per  acre? 

Ans.  8690. 

86.  What  cost  24  stoves,  at  815. 87^  apiece?         Ans.  8381. 

Remark. — The  preceding  examples  in  practice,  are  confined  to  Federal  Mon- 
ey; we  now  design  to  show  its  application  to  compound  numbers. 

87.  What  cost  12  acres,  40  sq.  rds.  of  land  at  820  per  acre? 

Ans.  8245. 

40  sq,  rds.  =  ^  of  an  acre.  Then  820  X  12J-  =  8225.     Hence— 

Operation,  To  find  the  value  of  a   quantity 

820  consisting   of    several    denomina- 

12  tions : 


8240=cost  of  1 2  acres. 
5= cost  of  40  sq.  rds. 


Ans. 


8245=costof  the  whole. 


Obs.  7.  Multiply  the  price  hy  the  number  of  the  denomination  at 
which  the  price  is  rated,  and  take  aliquot  parts,  to  find  the  value  of  the 
lower  denomincdlons.  The  sum  of  the  several  values  will  be  the  value 
of  the  entire  quantity. 

A  few  of  the  principal  aliquot  parts  in  compound  numbers,  are 
shown  in  the  following 


How  do  we  find  the  value  of  a  quantity  consisting  of  several  denomina- 
tions? 


210 


COMMON  ARITHMETIC. 


Sect.  XI 


TABLE, 


LONG    MEASURE. 

SQUARE    MEASURE, 

2  furlongs 

are 

i-  of  a  mile. 

40  sq. 

rds.    are   \  of  an  a( 

4 

u 

i-     - 

80 

2" 

6 

t( 

f =i+V  " 

120 

i(                   iC        3         1  _I_1        « 

4  —  2-r4 

40  rds. 

(( 

-*-    of  a  mile. 

AVOIRDirPOIS    WEIGHT. 

80  - 

*' 

jL      (t          (( 

6  cwt. 

is  ^  of  a  ton. 

160  rds. 

<( 

1      << 

2 

10      '' 

is  \     "     " 

240     " 

« 

4  —  2^^4 

15       '* 

is  l=\+\  ton. 

20     *' 

t( 

~  a  furlong. 

4  oz. 

is  ^  of  a  pound. 

1   ft. 

is 

i  of  a  yard. 

8  oz. 

is  \     '' 

6  inches 

(( 

1  a  foot. 

12  oz. 

is  i=i+L  ^' 

Tliese  are  the  parts  that  occur  the  most  frequently  in  common  busi- 
ness. Others  might  be  given,  but  it  is  thought  unnecessary,  as  the 
pupil  can  easily  discover  them  by  the  following- directions: 

Obs.  8.  Malce  ike  given  number  the  mimeraior  of  a  fraction,  and 
that  numher  which  it  takes  of  this  denomination  to  make  1  of  the  next 
higher,  the  denominator. 

Thus,  if  it  were  required  to  find  what  pait  of  an  hour  were  30 
minutes,  we  make  30  the  numerator,  and  60  (minutes  in  an  houj) 
the  denominator  ;  thus,  |J  =  \. 

of  wheat  cost,  at 
Ans.  $15.78i. 


88.  Plow  much   would  12 
a  1.25  per  bushel? 

Operation. 
^1.25 
12 


bu.  2  pks.  4  qts. 


81 

5.00 

=  cost  of  1 

2  bu. 

.62^ 

=  cost  of  2 

pks. 

.151 

=  cost  of  4 

qts: 

2  pks.  =  Y  a  bushel.  Hav- 
ing found  the  cost  of  12  bush- 
els, we  find  the  cost  of  2  pks., 
or  ^  a  bushel,  by  taking  ~  the 
price  of  1  bushel,  ($1.25.)  4 
qts.  =  ^  a  peck,  or  J^  of  2  pks. 
Therefore,  we  take  \  of  the 
15.78|  =  cost  of  the  whole,     cost  of  2  pks.,  (^0.62|)  as  the 

cost  of  4  qts.,  and  the  sum  of  all  these  is   equal  to  the  whole. — 

(Sect.  IV.   Art.  4.  Obs  4.  Rem.  2.) 

89.  How  much  will  15  bu.  3  pks.  6  qt  .  of  wheat  cost,  at  $1 .44 
per  bushel?  Ans.  ^22.95. 

The  learner  will  perceive  that  as  many  decimals  are  pointed  off 
in  the  result,  as    there  are  in    the  given  price  of  the  single  article. 


Nam«    some   of  the   aliquot   parts    of  Long   Measure?     Square   Measure? 
Avoirdupois   weight?     How  do  we   find  what  part  one  numher  is  of  another? 
In  finding  the  quantity  of  several  denominations,  how  many  decimals  do  we    •. 
point  off  in  the  result? 


Art.  2]  ANAiA'sis.  211 

90.  If  a  man  walk  24  miles,  80  rods  per  day,  fiow  far  can  he 
walk  in  48  days?  ,  Ans.  1164  miles, 

91.  How  much  would  15  acres,  2  R.  20  sq.  rds.  of  land  cost,  at 
88.50  per  acre?  Ans.  ^132. 81i^. 

92.  How  much  would  18  A.  1  R.  10  sq.  rds.  of  land  cost,  at  610 
an  acre?  Ans.  8183. 12f 

93.  How  much  would  32  A.  3  R.  30  sq.  rds.  of  land  cost,  at  ^15 
an  acre?  Ans.  $494. 06^. 

94.  How  much  would  6  T.  10  cwt.  20  lbs.  of  hay  cost,  at  ^8  a 
ton?  Ans.   852.08. 

95.  How  much  would  15  cwt.  2  qrs,  12  lbs.  8  oz.  of  cheese  cost, 
at$a2.50per  cwt.?  Ans.  ^195.31^.  _^ 

96.  If  a  man  work  12  hrs.  30  min.  per  day,  how  mucli  time  will 
he  work  in  36  days?  'Ans.   18 d.   18  hrs. 

97.  If  a  man  earn  $1 .50  per  day,  how  much  will  he  earn  in  24 
days,  18  hrs.  45  min.'?  "  Ans,  837.1 7f\. 

98.  How  much  would  157C8  lbs.  of  hay  cost,  at  5^8.40  per  ton? 

Ans.  v66.25. 

Solution. — A  ton  is  2000  lbs.  Therefore,  tlic  hny  costs  ^8 .  40  ~ 
2  =  $4. 20  per  thousand  pounds,  and  its  cost  is  found  according  to 
Sect.  VIII.  Art.  IK  Obs.  1.  to  be  ^^66.225.     Hence— 

To  find  the  cost  of  articles  by  the  ton,  or  2000  lb-,: 

Obs.  9.  Mtdtiply  half  the  cost  of  a  ton  by  the  iiumler  of  pounds, 
and  point  off  three  additional  dccimaU  from  the  right  hand. 

99.  How  much  will  be  the  storage  on  27480  lbs.  of  goods,  at 
S3. 60  per  ton?  Ans.  ^49.^64. 

100.  How  much  would  7240  lbs.  of  coal  cost,  at  ^:6.90  ptr  ton? 

Ans.  S24.978. 

101.  How  much  would  be  the  transportation  of  12415  lbs.  of 
goods  at  811.80  per  ton?  Ans.  $73,248. 

102.  At  87  per  ton,  how  much  would  1694  lbs.  of  hay  cost? 

Aus.  85.929. 

103.  A  farmer  sold  50  bushels  of  wheat,  at  81.25  per  bu.shel, 
and  took  his  pay  in  cloth,  at  80. 75  per  yard.  How  many  yards  did 
it  take?  Ans.  83i. 

Solution. — We  first  find  the  cost  of  the  wb^at  to  be  862.50. 
(Obs.  6-)  Next  we  divide  this  by  .75,  because  the  coth  is  80.75 
per  yard,  and  he  can  evidently  buv  as  many  yards  as  .75  is  con- 
tained in  862.50.  (Sect.  VI.  Art.  1.  Obs.  24.  a.) 


How  do  we  find  the  cost  of  articles  by  the  ton  of  2000  lbs.?     Detnoastrute 
this  rule. 


212  CORIMON    ARITHMETIC.  Scct,    XI 

Questions  of  this  kind  are  sometimes  solved  by  a  rule  calied  Bar- 
ter. » 

Barter  signifies  an  exchange  of  different  commodities,  at  ijrices 
agreed  upon  by  the  parties. 

Note. — It  is  not  necessary  to  give  aa  additioui^l  Rule  for  such  questions. 
Tliey  can  all  be  solved  by  Analysis^  a«d  the  process  is  so  simple  that  any 
scholar  of  moderate  capacity  can  easily  understand  it. 

104.  How  many  bushels  of  oats,  at  80.31^  a  bushel,  mu?t  be 
given  in  exchange  for  48  yards  of  muslin,  at  12^  cts.  a  yard? 

Ans.   19^. 

105.  How  many  cords  of  wood,  at  82.75  a  cord,  must  be  mven 
for  75  lbs  of  sugar,  at  lOj-  cts.  per  pound?  Ans.  2jj. 

106.  How  many  pounds  of  butter,  at  80. 16  per  pound,  must  be 
given  for  48  yards  of  calico,  at  80.25  jeryard?  Ans.  75. 

107.  If  I  give  45  acres  of  land,  worth  825  an  acre,  for  15  horses, 
how  much  do  the  horses  cost  me  apiece?  Ans.  $75. 

108.  Bought  a  hogshead  of  molasses,  at  80.45  per  gallon,  and 
paid  for  it  in  butter,  at  80. 18  per  pound.  How  many  pounds  did 
it  take?  Ans.   157^. 

109.  How  many  sheep,  at  81 .  75  apiece,  must  I  give  for  35  cows, 
at  810  apiece?  Ans.  200. 

1 10.  How  much  tobacco,  at  \2\  cents  a  pound,  must  I  give  in 
exchange  for  300  pounds  of  saleratus,  at  5  cts.  a  pound? 

Ans.   120  lbs. 

111.  How  many  pair  of  shoes,  at  82  .  25  a  pair,  will  it  take  to  pay 
for  144  yards  of  satmet,  at  8  1  .  25  a  yard?  Ans.  80. 

112.  Bought,  720  bushels  of  apples,  at  80.75  a  bushel,  and  paid 
for  them  with  1620  bushels  of  potatoes.  How  much  were  the  po- 
tatoes per  bushel?  Ans.  80.331, 

113.  A  man  mixed  5  lbs.  of  tea,  worth  80,50  per  pound,  with  6 
lbs.  worth  80.75  per  pound,  and  8  lbs  worth  80.80  per  pound. 
How  much  was  a  pound  of  the  mixture  worth?     Ans,  80.70jJ. 

Operation. 

80,50  X  5  =  82.50  =  the  price  of  5  lbs.  at  80,60  per  pound. 
0.75  X  6  =     4.50  =       *'         '*     6  lbs.  at  80.75 
0,80  X  8  =    6,40  =       "         <*     8  lbs.  at   88.80 


19  )   13.40  (,701^  cts.  Ans. 
13.3 

10 
The  whole  number  of  pounds  mixed  are  5  -f-  6  -J-  8  =  19  lbs. 

What  does  barter  sifinif  y  ? 


Art.  2.  ANALYSIS.  213 

The  cost  of  the  whole  mixture  is  $2.60  +  ^4. 60  +  $6.40  = 
Si 3. 40.  Then  one  pound  must  be  worth  $13.40  -f-  19  == 
$0,701  J.     Hence— 

To  find  the  cost  of  a  pound,  bushel,  <fec.,  of  a  mixture  : 

Obs.  10.  Divide  the  whole  cost  of  the  mixture  hy  the  whole  number 
of  simples. 

Obs  1 1 .  The  mixing  of  several  simples  of  differemt  qualities  to- 
getlur,  as  in  the  last  example,  to  form  a  comjJotmd  of  a  mean  or  mid- 
dle quality.)  is  called  Alligation. 

Alligation  is  of  two  kinds — Medial  and  Alternate. 

Obs.  12.  Alligation  Medial  i^  UMd  when  the  prices  and  quan- 
tities of  the  simples  are  given,  to  find  the  price  of  the  mixture  com- 
pounded of  them,  as  in  the  last  example. 

Remark. — Alligation  Alternate  is  not  much  used  iu  common  business,  and  is 
not  treated  of  in  this  work.  Its  chief  object  is  to  find  the  proportional  quan- 
tity to  be  taken  of  eacii  simple,  when  tlie  price  of  the  mixture,  and  of __the 
several  simples,  is giv«n. 

114.  A  farmer  made  a  mixture,  containing  12  bushels  of  oats, 
worth  $0-25  per  bushel  ;  15  bushels  of  corn,  at  $0.37l  per  bush- 
el,, and  8  bushels  of  peas  at  $0.62l  cents  per  bushel.  What  is  a 
bushel  of  the  mixture  worth?  Ans.  $0.38yJ. 

115.  A  grocer  mixed  30  gallons  of  water  with  60  gallons  of 
whiskey,  worth  $0.20  per  gallon.  How  much  is  a  gallon  of  the 
mixture  worth?  Ans.  $0.1 3i. 

116.  A  grocer  mixed  18  lbs.  of  sugar,  at  8  cts.  per  pound,  with 
22  lbs.,  at  10  cts  ;  24  lbs.,  at  12  cts.  ;  and  30  lbs.,  at  14  cts.  per 
pound.     How  much  is  a  pound  of  the  mixture  worth? 

Ans.  $0.1114. 

117.  A  man  mixed  40  gallons  of  wine,  worth  $2.25  per  gallon, 
with  30  gallons  of  brandy,  worth  $3.50  per  gallon,  and  20  gallons 
of  water.     How  much  was  a  gallon  of  the  mixture  worth? 

Ans.  $2.16|. 

118.  A  goldsmith  mixed  8,oz.  of  gold,  16  carats  fine  ;  12  oz.,  18 
carats  fine  ;  16  oz.,  20  carats  fine  ;  20  oz.,  22  carats  fine  ;  and  24 
oz.,  24  carats  fine.     Required — the  fineness  of  the  mixture. 

Ans.  21  carats. 

119.  James  and  William  wish  to  divide  24  cents  between  them, 
in  such  a  manner  that  James  may  have  3  as  many  times  as  Wil- 
liam.    How  many  must  each  one  have? 

Ans.  James  18,  and  WiUiam  6. 


How  do  we  find  the  eost  of  a  pound,  bushel,  &c.,  of  a  mixture?    What  is 
Alligation?    How  is  it  divided?    When  is  Alligation  Medial  usedt 


214  COMMON    ARITHMETIC.  Scct.   XI 

Solution. — Whilst  James  takes  3  cents,  William  takes  1  ;  and 
both  take  4  cents.  Then  James  must  rec  ive  as  many  times  3 
cents,  and  Wiiliam  as  many  times  1  cent,  as  4  is  contained  in  24. 
24  —  4  =:  6.  Therefore,  3X6  =  18  cents  for  James'  share,  and 
1  X  G  =  G  cents  for  William's  share. 

Proof.— 1 8  +  6  =  24.     18  =  6X3. 

Obs.  13.  The  process  of  dividing  a  number  into  two  or  more  parts 
that  shall  bear  a  certain  relation  to  each  other,  as  in  the  last  example,  is 
called  Proportional  Division. 

To  divide  a  number  into  proportional  parts  : 

Obs.  14.  Divide  the  number  to  be  divided  by  the  sum  of  the  parts : 
(lie  result  will  be  one  part,  from  which  the  other  parts  can  be  found. 

Proof  — Add  the  severed  results  together;  if  their  sum  is  equal  to 
the  number  divided,  tho  ivork  is  correct. 

Remark  1. — The  reason  of  the  proof  depends  upon  the  self-evident  princi- 
ple, that  the  whole  is  equal  to  the  sum  of  all  it^  parts. 

120.  Suppose  t^o  men  start,  one  from  Columbus,  and  the  other 
from  Wooster,  Ohio,  and  walk  towards  each  other,  the  former  at  the 
rate  of  3  miles  per  liour,  and  the  latter  at  the  rate  of  4  miles^per 
hour.  How  far  would  each  one  walk  before  ihey  would  meet,  the 
distance  being  85  miles? 

Ans.    j'he  former  ^Yalked  36^-  miles,  the  latter  474-  miles. 

121.  Divide  126  into  three  parts,  that  shall  be  to  each  oth>  r  as 
the  numbers  2,  3,  and  4.  Ans.  28,  42,  56, 

122.  Divide  144  ii-ito  4  parts,  the  numbers  of  which  shall  be  to 
each  other  as  the  number  3,  4,  S,  a'  d  6. 

Ans.  24,  32,  40,  48. 

123.  Divide  120  into  four  parts,  that  shall  be  to  each  other  as 
the  fractions  \,  ^,  \,  and  j.  Ans.  48,  32,  24,  16. 

Remark  2. — By  multiplying  each  fraction  by  the  least  common  multiple  of 
their  deaominators,  we  find  they  are  to  ejch  other  as  the  numbers  6,  4,  3, 
and  2. 

124.  The  standard  for  gold  and  silver  coins  in  the  United  States 
is  9  parts  pure  metal  to  1  part  alloy.  How  much  of  each  is  there 
in  an  eagle,  which  weighs  10  pwts.  18  grs.? 

Ans.  Gold,  9  pwts.  16j  i^rs.  Alloy,  1  pwt.  ly  grs. 

125.  Gunpowder  is  composed  of  76  parts  of  nitre,  14  of  char- 

What  is  Proportional  Division?  How  do  wp  divide  a  number  into  propor- 
tional parts?  What  is  the  method  of  proof?  Upon  what  does  the  reason  of 
this  method  of  proof  depoud?  How  do  we  find  the  relation  which  fractions 
bear  to  each  ather? 


Art.    2*  ANALYSIS.  215 

coal,  and  10  of  sulphur.     How  much  of  each  is  there  m   200  Ibs.l 
Ans.  Nitre,  152  lbs.   Charcoal,  28  lbs.  Sulphur,  20  lbs. 
126.  If  a  man  ride  288  miles  in  8  days,  riding  9  hours  per  day, 
how  far,  at  the  same  rate,  can  he  ride  in  15  days,   riding  12  hours 
per  day? 

Sohition. — If  he  rides  288  miles  in  8  day?,  he  will  ride  288  -—  8 
=  36  miles  per  day  ;  and  if  he  rides  but  9  hours  per  day,  he  will 
ride  36-J-9  =:  4  miles  per  hour.  Again — if  he  rides  4  miles  per 
hour,  he  will  ride  4  X  12  =  48  miles  in  a  day  of  12  hours  length  ; 
and  in  15  such  days  he  would  ride  48  X  15  =  720  miles. 

Ans.  720  miles. 

Or,  the  operation  may  be  performed  by  cancelation.     Thus  : 

The  learner  will  perceive  that  we  multiply 
and  divide  the  same  as  in  the  above  solution. 


15 

12                      I 

Ans. 

720  miles. 

Note. — Questions  of  this  kind  are  usually  solved  by  Compound  Proportion, 
which  is  explained  in  the  next  section.  But  the  solution  by  Analysis  is  prefer- 
able to  any  other.  Tlie  learner  will  observe  that  the  canceling  part  is  not 
the  solution,  but  merely  the  operation.  The  operation  is  the  mechanical,  and 
the  solution  the  mental,  or  reasoning  part,  of  the  work,  aud  a  pupil  ouglit 
never  to  accustom  himself  to  the  use  of  the  former,  without  the  aid  of  the 
latter,  as  a  thorough,  knowledore  of  all  the  reasons  why  he  performs  his  operations 
an  he  does,  is  absolutely  necessary,  if  he  ever  wishes  to  become  a  goud  arithme- 
tician. ^ 

127.  If  a  man  plouglr*?^,  acres  in  10  days,  how  long  will  it  take 
18  men  to  plough  230  acres?  Ans.   10|j  days. 

128.  If  5  men  earn  ^75  in  12  days,  how  long  will  it  take  8  men 
to  earn  S200?  Ans.  20  days. 

129.  If  15  men  dig  360  rods  of  ditch  in  4  days,  how  many  rods 
will  24  men  dig  in  12  days?  Ans.   1728. 

130.  If  12  horses  eat  126  bushels  of  oats  in  14  days,  how  many 
bushels  will  16  horses  eat  in  24  days?  Ans.  288. 

131.  If  a  family  of  5  persons  spend  $87.50  in  7  months,  how 
much  will  they  spend  in  16  months,  if  6  more  persons  are  added  to 
the  family?  Ans.  ^440. 

132.  A  farmer  has  i-  of  his  hogs  in  one  field,  ^  in  another,  and 
24  in  a  third  field.     How  many  has  he  in  all?  Ans.   144. 

Solution. — i-  -[- 1  =  I  =  what  he  has  in  two  fields  ;  then  the 
third  field  must  contain  J  —  |  =  J-  of  the  whole.     But  the  third 

Explain  the  difference  between  the  operation  and  the  solution  of  a  question 
in  mathematics?  Should  the  former  ever  be  used  without  the  latter?  Wbv 
not? 


216  COMMON    ARITHMETIC.  Scct"'  XI 

field  contains  24;  therefore,  24  is  i-  of  the  whole,  and  24  X  6  = 
144,  the  whole  lot. 

Note. — Questions  of  this  kind  are  sometimes  solved  by  a  rule  called  Posi- 
tion, or  Trial  and  Error.     Analytic  solutions,  however,  are  preferable. 

Remark  3. — Position  might,  with  propriety,  be  called  the  Guess  Work 
Rule.  In  Arithmetic,  all  qu.-'stions  should  be  solved  upon  true  principles, -fyfith' 
out  any  guess  work  about  it  ;  and  aa  all  examples  in  Position  can  be  solved  by 
Analysis,  to  give  an  extra  rule  for  their  solatiou  would  be  unnecessary. 

133.  In  a  certain  school  j  of  the  pupils  study  Arithmetic,  f  study 
Philosophy,  12  study  Algebra,  and  the  remainder,  who  are  7  of  the 
school,  study  various  branches.  How  many  scholars  are  therein 
the  school?  Ans.  84. 

134.  or  a  certain  army,  \  were  killed,  \  were  wounded,  ~  were 
taken  prisoners,  and  3000  fled.  How  many  were  in  the  army  at 
first?  Ans.   12000. 

135.  A  drover  has  sheep  in  6  lots  :  In  the  first  he  has  \  of  his 
flock ;  in  the  second,  j ;  in  the  third,  i- ;  in  the  fourth,  ~  ;  in  the 
fifth,  — ;  and  in  the  sixth  he  has  84.     How  many  sheep  has  he? 

Ans,  480. 

136.  If  a  pole  is  \  in  the  mud,  i  in  the  water,  and  20  feet  out  of 
the  water,  what  is  its  length?  Ans.  48  feet. 

137.  A  schoolmaster  being  asked  how  many  children  he  had,  an- 
swered :  "  If  I  had  as  many  more  as  I  now  have,  |  as  many,  ~  as 
many,  j  as  many,  and  ~  as  many,  I  should  then  have  270."  How 
many  had  he?  Ans,  75. 

138.  What  number  is  that,  ~,  \,  and  i-  of  which  is  36? 

Ans.  48. 
139-  What  number  is  that,  ~  of  which  exceeds  ~  of  it  by  10? 

Ans.  60. 

140.  What  number  is  that,  from  which  if  you  take  J,  i-,  and  ~  of 
itself,  the  remainder  will  be  44?  Ans.  96. 

141.  What  number  is  that,  which,  if  you  increase  it  by  ~,  yV» 
and  y2  <^f  itsilf,  the  result  will  be  157?  Ans.   120. 

142.  A.'s  age  is  3  times  B.'s,  and  twice  A.'s  age  equals  C.'s  ; 
the  sum  of  all  their  ages  is  80  years.     Row  old  is  each? 

Ans.  A.  24,  B.  8.  and  C.  48  years. 

This  example  might  be  performed  by  Proportional  Division,  B.'s 
part  being  1,  A.'s  part  3,  and  C.'s  3  X  2  =  6  ;  then  6  +  3  +  1  = 
10,  the  sum  of  all  the  parts. 

What  might  Position  with  propriety  be  called?  How  shoi^-ld  all  questions  in 
arithmetic  be  solved?  What  do  you  mean  by  true  principles?  How  can  all 
questions  in  Position  be  solved? 

tBv  trueprintiples,  are  meant  principlee  Capable  of  demonrtr»tion» or  pft)or» 


Art.  ^.  ^  A^iAL^stRi  in 


.ft- 


143.  A.,  B..  and  0.  talk  of  their  ages.  A.  said  his  age  was  Y 
of  B.'s,  and  C.  said  his  age  was  J  pf  the  sum  of  the  ages  of  A.  and 
B.  both  ;  the  sum  of  all  their  ages  was  121  years..  Required — the 
•age  of  each. 

V  Ans.  A.'s  age,  30  years.  B.'s,  1.4  years.  C.'s  77  years. 

144.  Am  n  bought  a  sheep,  cow,  and  horse;  the  cow  cost  5  times 
as  much  as  the  sheep,  and  the  horse  cost  5  times  as  much  as  the 
cow  and  sheep  both.  They  all  cost  ^72.  Required^ — ^the  cost  of 
each? 

Ans.  The  sheep,  82  ;  the  cow,  810  ;  and  the  horse,  860. 

145.  A.  and  B-  have  the  same  annual  income.  A.  saves  ^  of 
his,  but  B.,  by  spending  ^  as  much  again  as  A  ,  loses  875  in  the 
course  of  the  year.  Required — the  amount  of  the  annual  income 
of  each.  -  Ans.  8300. 

146.  A.  and  B.  have  the  same  annual  income  :  A.  saves  yV  of 
his,  but  B.,  by  spending  850  per  year  more  than  A.,  at  the  end  of 
5  years  finds  himself  875  in  debt.  Required — the  annual  income 
of  each.  Ans.  8350. 

Solniion. — B.  runs  in  debt  375  in  5  years.  This  is  875-^5  = 
815  a  year  more  than  his  income  ;  and  as  he  spends  850  per  year 
more  than  A.,  A.  must  save  850  —  815  =  835  a  year ;  and  as  A. 
saves  y\  of  his  income,  their  income  must  be  835  X  10  =  8350. 

147.  A  men  hired  90  days  on  .these  conditions  :  for  every  day  he 
worked,  he  was  to  receive  8Q.75  ;  and  for  every  day  he  was  idle, 
he  was  to,  pay  80 .  25  for  his  board.  At  the  end  of  his  time  he  re- 
ceived 840.  How  many  days  did  he  work,  ynd  how  many  days 
was  he  idle?  Ans,  He  worked  62^^  days,  and  was  idle  21\. 

148.  A.  and  B.  commence  trading  with  equal  sums  of  money  : 
A.  gained  a  sum  equal  to  \  of  his  stock,  and  B.  lost  8200.  B.  had. 
then  I  as  much  money  as  A.     How  much  had  each  at  first? 

'f'^o   :''  Ans... 8 1200::': 

149.  A  man  left  his  property  to  his  3  children,  gloving  A.  L  Tfant- 
ing  8200  ;  to  B.  |,  and  to  C.  the  rest,  which  was  8200  less  than 
the  share  of  B.  What  was  the  value  of  the  estate,  and  what  was^ 
each  one's  share?  '•      - 

Ans.  Estate,  83000  ;  A.'s  share,  8800;  B.*s  share,  81200;  and  C.'s. 

[share,  81000. 

1 50.  A  person  being  asked  the  time,  replied :  "  The  time  past,* 
noon  is  equal  to  —  of -the  time  till  midnight."     What  time  was  it,? 

/t--;  Ans.  45  min.  past  3  o'clock. 

^  151.  'A  certain  pole  is  composed  of  three  pieces  :  The  top  piec^ 
is  15  feet  long,  the  middle  piece  is  as  long  as  the  top  piece,  and  ^  nf^ 
the  length  of  the  lower -piece ;  and  the  lower  piee**  i»  as 


218  COMIVION    ARITHMETIC.  ScCt.    XI 

both  the  oth  r  pieces.     How  long  is  the  pole  and  how  long  is  each 

.piece  i 

'Ans.  Length  of  the  p  le,  100  feet;  length  of  pieces,  15,  35,  60  ft 

152.  A  man  has  a  gold  and  a  silver  watch,  and  a  chain  worth 
■^20.  If  he  put  the  chain  on  the  gold  watch,  its  value  Avill  be  6 
times  that  of  the  silver  watch  ;  but  f  he  put  the  chain  on  the  sik-er 
;vv'atch,  its  value  will  be  but  half  that  of  the  gold  watch.  Required 
— -the  y'4l]ie  of  the  two  watches. 

,,;^v;    .     J  Ans.  The  gold  one  ^70;  the  silver  one  8 15. 

Solidion.—rKs  tli^  chain,  when  put  on  tbe  gold  watch,  makes  its 
value  6  times  that  of  the  silver  watch,  the  silver  watch  must  be 
worth  $20-^-  6  =  S3i-  J^\  *  f  the  value  of  the  gold  watch.  Ther^r- 
fore  when  the  chain  is  put  on  the  silver  watch  its  value  must  be  «f  20 
-(-  'f  3J  -f-  i  of  the  y^lye  of  the  gold  watch.  But  by  the  question 
this  is  -2*  the  value  of  the  gold  watch  ;  therefore,  twice  this,  or  ^40 
-\-  ^6|  -\t  \  of  the  value  of  tjie  gohi  watch  must  be  the  value  of  the 
gold  walch.  Now  J$40  +  v6|  =  '1i>46|  ;  and  from  the  above  reason- 
ing, this  must  be  |  of  the  value  of  the  gold  w^atch.  If  $46|  is  |, 
$46|-j-  2  =  i23i-,  is  J  and  ^^  X  3  s=:  $70,  the  value  of  the 
gold  watch. 

153.  A  man  has  two  horses,  and  a  saddle  worth  825.  If  he 
puts  the  saddle  on  the  first  horse,  his  value  will  be  y  of  that  of  the 
^cond  horse ;  but  if  he  puts  the  saddle  on  the  second  horse,  his 
T*^lue  will  be  J  of  that  of  the  first  horse.  Required — the  value  of 
each  horse.  Ans.  The  first  $60  ;  the  second  845. 

'  154,  The  hou"  and  minute  hand  of  a  clock  are  together  at  12 
o'clock.     At  what  time  are  they  next  together? 

••  Ans,  5  min,  27jy  s-c.  past  1. 

Solution. — The  minute  hand  moves  over  12  spaces,  whilst  the 
hour  hand  moves  over  but  1,  and  therefore  gains  11  spaces  in  an 
hour.  If  the  minute  hand  overtook  the  hour  hand  precisely  at  1 
^o'clock,  it  would  gain  12  spaces  in  an  hour.  But  it  gains  only  11 
spaces  \n  an  hour;  consequently  to  overtake  the  hour  hand  it  must 
ga  n  i-y  of  the  distance  it  has  to  run.  \j  of  1  hour,  or  60  minutes 
is  \  iiQiir,  ,5  minutes,  27yy  seconds.  (Sect.  IX.  Art  3.  Case  2.) 

156,  At  ^hat  time  between  3  and  4  are  the  hands  of  a  clock  to- 
gether? Ans.   \Q  min.  21/y  sec.  past  3  o'clock. 

j1,56.  At  ^^%t  tipae  between  10  and  11  are  the  hands  of  a  clock 
ItQgethe^?  Ans.  54  min.  32yY  sec.  past  10  o'clock. 

157.  At  what  tiijjte  j^etwe^en  9  and  10  are  the  hands  of  a  clock 
exactly  opposite  each  ot^er?  Ans.   16y,  min.  past  9  o'clock. 

Suggestion. — The  learne^  wiH  perceive  that  to  be  opposite  each 
^ptj^^ef .  the  hands  must  be  30  xpiiiit^e^  apa^t.    Now  at  9  o'clock  th^ 


Ar^  1.  '^^'^^^^^I^MP         ^  ^^'^^ 

minute  hand  is  15  minutes  aliead  of  tlieliour  liand  ;  and  tlierefore, 
at  has  but  15  minutes  more  to  gain  to  be  opposite  the  hour  hand. 

1 58.  At  what  time  bet\voen  2  and  3  are  'the  iiands  «©f  a  eleek  ex- 
actly opposite  each  other?  Ans.  43y\  min.  past  2  o'clock. 

159.  Four  men  trade  together  :  A.  put  in  yy  as  much  as  B.  ; 
G.  put  in  y|  of  II  of  what  A.  and  B.  both  put  in  ;  and  D.  ,put  in 
_7_  of  1|  of  what  B.  and  C.  put  in.  They  all  put  in  t9875.  What 
was -each  one's  share? 

Ans.  A,'s  ^1750  ;  K's  $£750  ;  C.'s  $3500  ;  and  P.'s  $1:875, 

160.  A.,  B.  and  C.  trade  in  company:  A.  put  m  $4500-  J  of 
what  A.  put  in  was  equal  to  J  of  what  B.  put  in  ;  and  -~  of  what 
B.  put  in,  added  to  yj  of  what  A.  put  in,  was  y-  of  what  C.  put  in. 
They  gained  a  certain  sum,  of  •which  A-.'s  share  v/as  equal  to  4  of 
what  B.  and  C.  put  in  ;  B.'s  share  was  equal  to  ~  of  what  A.  :>nd 
C  put  in.;  and  C.'s  share  was  equal  to  \\  ^^  what  they  all  put  in. 
What  wa>  the  whole  amount  invested?  How  much  did  B.  and  C. 
put  in  ;  what  did  they  gain,  and  what  was  each  one's  -sliare  of  the 
gain? 

C  Whole  stock,  $15000;  B.  put  in  $5000  ;  C.  |)ut  in  $5500. 
Ans.    <  Whole  gain,   $^0000  ;  A.'s  share,    SSOOO ;  R.'s   share, 
(      $3333.331 ;  C.'s  share.  $3666. 66f. 


SECTION  XII. 

KATIO  AND  PROPORTIOX. 

Article  1.     Ratio. 

Obs.  1.  Ratio  me«w5  relation,  and  signifies  i^mt  rdation  which 
-one  nuniher  has  to  another  of  the  same  kind.  It  is  expressed  hy  the 
■quotient  of  one  divided  hy  the  other. 

Thus,  if  we  inquire,  What  is  the  axttio  of  A  to  9.1  if  we  make  the 
€rst  number  (4,)  the  standard  of  comparison,  the  question  resolves 
itself  into  this  :    What  part  of  ^  is  11  and  the  answer  is,  2  is  |-  of  4. 

Again  :  if  we  make  a  second  number  (2,)  the  standard  of  com- 
parison, the  question  is.  What  part  of  Us  ^1  and  the  answer  is,  4 
is  twice  2. 

The  first  of  these  methods  is  called  the  Frinch  method,  by  which 
the  ratio- V  4to2is2-i-4  =  ^. 

What  does  ratio  mean?  What  does  it  signify?  How  is  it  expregsed?  How 
do  we  find  the  ratio  between  two  numbers? 


MO^  ^^         ccaaioN  arithmetic.  Sect.  XII' 


CCt|AI< 

d  i^Pcalle 


Tlie  second  method  ilPcalled  the  English  method,  by  which  the 
?.ntio  of  4  to  .2  is,  4-f-  2"=r£. 

Each  method  has  adYantages^pecTiliar  to  itself,  but  we  shall  follow 
iHe  i^?w2c A  method,  as  it  is  generally  adopted  in- this  country.. 
.Hence — 

To  find'  the  ratio  of  chc  Rumber  to  another  r 
Qbs.  2.     Divide  the  second  numher  by  the  first,. 

MENTAL    EXERCISES, 

r. .  What  is  the  riitio  of  9  to  3?'  of  6  to  1 2?  Ans.  ^,  2-. 

2:  What  is  the  ratio  of  16  to-  21*  Of  21^  to  7t  Of  48  to-  12?: 
©f  63to  9?     Of  84  to  7? 

3-  What  is  the  ratio  of  30  to  6?  Of  36  to  4?  Of  120  to  lOt^ 
Of  132  toll? 

4.  What  is  the  ratio  of  3  to  41  Of  54o-8?  Of  7  to-9?  Of  8* 
t.\    5?     Of  23  to  37? 

5.  What  is  the  ratio  of  15  to  251  Of  18  to  36?  Of  36  to  84?' 
Of  90  to  100? 

6;  What  is  the  ratio^  of  75  to  25?  Of  144  to  48?  Of  132  to. 
11728? 

T.  What  is  the  ratio -of  288  to  676?     Of  132  to  96?     Of  l^e^to^ 

aio? 

Obs.  3.     The  two*  numbers  thus  compared,  when  spoken  of  to- 
gether, as  in  the  preceding  examples,   are  termed  a  couplet  ;  buti 
when  spoken  of  separately,  they  are  called  the  terms  oi  the  couplet; 
Obs.  4:     HhQ  first  terra  is  called  the  antecedent,  because  it  comes^ 
before  the  other  ;  the  second  term  i&  called  the  consequent,  because  itf 
JdUows  afUr  the  ot'lier. 

8.  Of  what  numbers  is  f  the  ratio?  Or,  what  number,  divided 
lay,  another,  will  produce  |? 

It  is  evident  that  the  least  numbers  thai"  will  do 'this  are  5  and  6v 
c^'.the  ratio  of  5  to  6  is  f.     Hence — 

To -find  two  numbers  of  which  a  given  fi'^ction'  is- the -ratio  : 

Obs.  5.-  Make  the  numerator  tJie  second -number,  and  the  denomina- 
tcrthe  first  term  of  the  couplet. 

9i   Of  what  two- numbers  are-the   following,-,  fractions  the  ratio?. 

2  ^.    7  ..   6  .     12.     15,     1_6        2^0^.     2  8^8^.     7_iL  •     ^JL"  •     \.1A^'     ^JJU  ? 
"S^F'T'      11'       9'       4*      25'      144'     63'       48    f     4370'        576     ' 

ID;.  Of  what  two  numbers  is  7  the  ratio? 

When  two  numbers  compared  are  spoken  of  togefber,  what  are  they  called?*- 
When  spoken  of  s^p^rately,  what  are  they  called?  What  is  the  first  terra 
cafcl^KJ?'  Why?  What  is  the  second  term  called?  Why?  How  do  we  fi,nd 
twaniumbersj.or  which  a  given  fraction  is  the  ratio? 


Art.    1.  RATIO   AND  TROPORTION.  J^P  ^2281 


It  is-evident  that  we  -wish  to  find  tw^lHDcrs,  one  of  \\^ic'lniiu^- 
tiplied  by  7  -will  produce  tbeotber.  1  %it  assuming-  l  as  one.nirat- 
ber,  the  other  must  be  1X7  =  7;  or,  assuming  2  as  one  nunibar, 
the  other  is  2  X  7  =  14  ;  also,  3  X  7^21,  5&c.  That  is,  ,7  is.the 
ratio  of  1  to  7;  of  2  to  14  ;  of  3  to  21,  &c.     Hence — 

To  find  two  numbers  of  which  a  given  integer  is  the  ratio  : 

Obs.  6.  Take  any  number  as  the  first  term  of  the  couplet /aiti, 
'.multiply' it  by  the  ratio  to  obtain  the  second  term  of  the  couplet.  rDi>e 
same  rule  will  apply  when  the  ratio  is  a  fraction,  and  other  terms 
are  desired  than  the  numerator  and  deiiOminator. 

11.  Of  what  two  numbers  is  6  the  ratio?  9;  4;  8;  12;  40;  l-S; 
26;  38;  18:? 

12.  Of  what  two  numbers  is  3  the  .TKitio?  5;  11;  13;  23;  ^; 
,108;   150? 

Re»a,xk  1. — As  the  ratio  is  found  by  dividing  cue  nomber  by  another,  itcaa 
always  be  expressed /rac^tonfl%.  Thus,  the  ratio  of  6  to  3  is  |-  ;-'a«d-the  ra- 
tio of  8  to  16  is  y. 

2.— Or,  the  ratio  may  be  expressed  by  a  couplet  of  dots  ( : )  placed  belvveaE 
•the  terms  of  the  couplet.  Tnus  the  ratio  of  10  \.t>  lo  may  be  eipies.sed"thu€^ 
10  :  15.     Hence— 

Obs.  7.  Ratio  may  be  expressed  in  two  ways,  and  it  isiimndterml 
-which  ive  use. 

Obs.  8.  As  ratio  expresses  the  relation  of  numbers,  owe  na??t5gr 
cannot  have  a  ratio  to  another  nwnberofu  different  kind.  Because  it 
would  be  absurd  to  c  jmpare  bushels  with  hours,  or  feet  withpounds, 
as  a  bushel  cannot  be  said  to  be  part  of  an  hour,  or  a  foot  longer  or 
shorter  than  a  pound.  But  we  can  compai-e  bushels  with  bushels,  ^or 
feet  with  fcet*or  .pounds  with  pounds,  because  they  express  things  .of 
,the  same  kind. 

Remark. — From  this  we  perceive  that  ratio  refers  only  to  the  relative,  anrS 
not  to  the  «6soZwfe  magnitude  of  the  two  quantities.  Fori  lb.  has  the  same 
relation  to  2  lbs.  as  1  ton  to  2  tons,  or  1  ounce  to  2  ounces. 

13.  What  is  the  ratio  ef  2^pecks  to  1  peck? 

Ans.  I,  or  1  peck  is "|  of  2  pecks. 

How  do  we  find  two  numbers  of  which  a  given  integer  is  the-ratio?  W.iU 
Ihe  same  rule  apply  when  the  ratio  is  a  fraction?  How  can  ratio  always  be 
expressed?  How  may  it.elherwise  be  eixpressed?  What  is  the  iufnirence  dt^ 
.duced  frem  this?  Can  ratio  express  the  relation  betA'een  numbers  of  diff.-renrt 
kinds?  Pan  we  coinnare  bushels  with  hours?  Why  not7  Can  we  oompure 
iie  t  with  ^louuds?  Why  not?  To  what  can  wl-  cumpare  bushels?  With  whal 
can  we  compare  fe-i?  VVitii  what  can  we  compare  pound*?  v\  by?  Does 
ratio  refer  to  the  relative  or  absolute  mi^gniiude  of  two  auantiti^s^  Giv^t 
«u  example. 


222  WJL,         COMMON    ARITHMETIC,  Scct,  XII 

14.  What  is  the  ratioii6f-^  bushels  to  3  pecks? 

IS^OTK. — First  reduce  both  to  tiie  same  denomination..    2  bu.  =  8  p^Rs.- 

Ans.  J. 

G'bs.  9.  Hence — When  the  terms  of  ike  couj^lei  are  of  the  same 
Mnd;  hut  of  dij%re7it  dtnominations,  they  must  both  he  redztced  to  the 
same  denomination  before  their  ratio  can  be  found. 

15.  What  is  the  ratio- of  2  feet  to  4  yards? 

16.  What  is  the  ratio  of  1'  ton  to-  m^  ewt.?  Of  3*  seconds  to  4 
yzMTiiites? 

17.  Wliat  is  the  ratio  of  120  sq..  rds.  to  2  acres?  Of  7  acres  to 
21^  acres?  Of  320  acres  to  1  sq.  milel  Of  6  miles  to  3  miles? 
Of  8  miles  to  4  furlongs ?> 

1.8.  What  is  the  ratio  of  2  miles  to  180  rods?  Of  1  bushel  to 
6^qts.?     Of  1  gallon  to  2  qts.?     Of  X  ton  to  1^  lbs.? 

Remark. —  As  the  ratio  of  any  two  numbers  can  always  be  expressed  frac- 
Tlr  n:;lly,  ii  is  obvious,  that  as  fur  as  the  vnhte  is  concerned,  the  same  general 
principles  are  r qually  applicable  to  both  ratio  and  fraGliaas.     He»ce — 

Obsi  IK).  Jf  we  multiply  the  consequent,  or  divide  the  antecedent  hy< 
any,  number,  we  midtijjly  the  ratio  by  that  number.  (Sect.  VIll.  Aat. 
±  Obs.  1  and  5.) 

Thiis^  Tlie  ratio  of  2  :  4  is  2. 

Multiply  ingUhe  consequent. by  2,  the  ralio  of  2  :  8  is  4,  or  2X.2. 
Dividing  the  antecedent        by  2,       "        "      i  :  4  is  4,  or  2X2- 

Rb^akk.' — This  is  evident  from  the  fact  i\vd%w»  either viidtiphj  the  numerator 
or  diviae  the  dfnomiriatvr,  eilht-r  of  which  has-  the  same  efSect  upon  tlie  quo- 
tient. (Sect.  Vlll.  Art.  2.  Obs.  7.) 

Obs.  \1\  If  we  divide  the  conseqzient,  or  muUijjly  the  antecedent  by 
any  number,  2ve  divide  the  ratio  by  that  nwnbcr.  (Sect.  VIll.  Art.  2. 
Obs.  2and4^_). 

Thus,  The  ratio  of  4  ;  16  is  4. 

Dividing  the  conseq,uent      by  2,       "         •'  4  :     8  is  2,  or  4-f-2. 
Multiplying  the  antecedent  by  2;.       "  '^  8  ;   1 6  is  2,  or  4^-2, 

Remark. — This  is  evident  from  the  fact,  thai  tee  (ither  divide,  the  nvmcrator 
«ff?nultipkj.ihe  defiominufor,  f'lihi'r  of  whi*?))  lii-s  iS;i>  sa-we  etlect  upon  the  quo-- 
ti^'nt.  (Sect.  Vlll.  Art.  2.  Obs.?:) 


ii  When  tha  terms  of  the  couplet  are  of  tiie  i?anie  kind,  but  of  different  de- 
»s«iinations,  how  do  we  proceed?  Tlie  ratio  of  any  two  numl^ers  can  always 
W  p»^|>r^ssed  fractionally  ;  what  is  the  rnferemre  deduced  from  this  fact?  What 
eS^ct  does  it  have  upon  the  ratio,  to  rnul'riity  the  coiisc-queiit,  or  divide  Ihean- 
te3e<'ent  by  anj  ninnber'T'  Bh»w  why  ^bis  is  cor'f  cl.  What  eftVc;  dots  it  have! 
upon  the  ratio  to  divide  the  consequeut,  or  multiply  the  antecedent,  by  any 
»atQber?     Show  why  this  is  correct  ? 


Art.    I.  B\TIO    AND   PROAORTON.  223 

Obs.  12.  If  we  multiply  and  d'vide  both  the  antecedent  and  the 
consequent  by  the  same  number,  we  do  not  alter  the  ratio.  (Sect  V,II. 
Art.  2.  Obs.  8.) 

Tbtrii.  Tlie  ratio  of  l-S  :     QU\. 

Makiplying  both  terms  of  the  coup' et  by  .%       -'  '*  36  :  ISisi^. 

Dividing*  both  term- of  the  couplet      by  3,       "  "4:     2is^. 

Remark. — This  is  evident  fron  th"  fiict,  thit  we  multiply  or  d'mile  both  ditti- 
sor  and  dividend  by  the  Sf/me  number,  which  does  not  alter  tne  quo'ieiit.  (.Sect. 
VI.  Art.  1.  Obs.  2d.) 

Obs.  13.     If  the  arUecedent  and  consequent  are  equal,  the  raJio  is^ 

1,  and  it  is  called  a  ratio  of  equediiy    (Sect.   VIII.   Art,   2.  Obs-'^ 
10.  a,) 

Thus,  the  ratio  of  3  :  3  is  1.  or  3  is  equal  to  3. 

Resiark. — "^'his  is  hvid'^nt  f mm  the  fa :t,  tint  the  ditnsor  and  diDiiendar^    r. 
equal.  (Sect.  VI.  Art.l.  Obs.  26.  d.)  '     -        '^ 

Obs.  14.  -7/-'  the  aixfe-edenl  is  g;re(tfer  than  the  comerracnt.  the  ratlo' 
is  leis  than  1,  and  It  is  called  oi  ratiij  of  imqitality.   {^.Secc.  Vill,  .irt.- 

2.  Obs.  10.) 

Remark.—  This  is  evident  from  the  fuo',  that  the  divisor  is  greater  than  the- 
dioiJend.  (Sect.  VI.  Art.  1.  Obs.  2&-  d.) 

Obs,  15.  If  the  anteceffent  is  less  than  the  consequent,  the  railo  is  • 
greater  than  1,  and  it  i»  called  a  ratio  of  qreater  in/^qual-tyL  \^(^%. . 
VIII.  Art,  2.  Obs.  10.  h.) 

Thus,  the  ratio  of  5  :  \h,  or  3,  is  a  ratio  of  gre£cte?r~i6equa]ity. . 

Remark. — This  is  p  vide  at  fron  the  fact,  that  the  dividend  is  grmtjfr.tJuin  the 
divisor.  (Sect.  VI.  Art.  1.  Oi)s.26.  d.) 

19.  Of  what  kind  is  the  ratio  of  5  to  7? 

Ans.  A  ratio  of  greater  inequality.. 

20.  Of  what  kind  are  the  ratios  of  the  frllowiag  couplets  :  6  :  4>; 
8:8;  7  :  14;  36  ;  24;  15  :  15?  20  ;  40;  60  :  30;  15  :  24;, 
23  :   17?  -'• 

Obs.  16.  Ratio  is  of  two  kinds — Simple  and  Compound. 


■T 


Whit  effect  does  it  have  upon  the  ratio  to  multiply  or  divid*^  both  the  ante-, 
cedent  iin<i  consequent  by  the  siinie  number?  Show  w  y  tWs  is  correct.  IC\ 
the  ant'^cedent  and  consequent  are  equ-^i,  wimt  is  the  ratio?  'i  hat  is  it  called? 
Why  is  this  correci?  If  ih-^  antecedent  is  ^reat^r  thm  ih-e-o.o-nsf'q  lent,  wliat 
\»  the  rilio?  What  is  it  called?  Why  i-*  this  cyrracl?  If  Hie  atitfced-^nt  is  lesa 
than  the  conseq*i**nt,  what  is  the  ratio?  Wiia.t  is  itca.Ued>?  Why  i&tliiscoj-. 
r«ct?     How  is  ratio  divided? 


224  COMMON    ARITHMETIC.  Sect.  XII 

a.  A  Simple  ratio  is  the  ratio  of  a  single  expression,  or  couplet. 

b.  A  Compound  ratio  is  the  ratio  of  the  products  of  tJie  corres- 
ponding terms  of  two  or  more  couplets. 

Thus,  the  ratio  of  6  ;  2,  or  12  :  24  is  a  simple  ratio.  . 
Again  the  simple  ratio  of  3  I  6  is  2. 

And  the  simple  ratio  of  2  :*  8  is  4. 

And  the  ratio  compounded  of  these  is  the  same  — — 

;as  the  simple  ratio  of  6  :  48,  which  is  8. 

Note.— Compound  ratio  is  not  different  from  any  other  ratio.  It  is  only 
used  in  some  cases  tp  denote  tJie  origin  of  the  ratio, 

21.  Whi*  is  tlie  simple  ratio  of  4  *  2?  Of  8  '  16'^  Of  48  '  2Q' 
Of  bQ  :  28?  *        '  *       ' 

22.  What  is  the  compound  ratio  of  the  following  ratios  : 

6  :  i2>        8  :    )        2:8)        12  :  6)        18  :    9) 

4  ;  24p        9:2^'        6  :  24^;         24  ;  8^;  12  :  .36V 

4  :  12^        16  :  9)         4  :  16) 

(  Article  2.     Proportion, 

'Obs.  1 .     A  Proportion  consists  of  an  equality  of  ratios. 

"^hus,  the  ratio  3  :  9  is  equal  to  the  ratio  of  4  ;  1 2;  that  is,  the 
ratk)of  both  is  3.  Hence,  the  two  couplets  3  :  9  and  4  ;  12  form 
a  proportion. 

Obs.  2.  The  terms  of  the  two  couplets  which  compose  the  propor- 
tion are  called  Proportionals. 

Obs.  3.  Proportion  is  usually  expressed  hy  four  dots  (  !  .*  ) 
placed  between  the  two  coup)lets. 

Thus:  4:8;  :  6  :  12  is  a  proportion,  and  is  read,  4  is  to  8  as 
6  is  to  12 ;  or,  the  ratio  of  4  to  8  is  the  same  cis  that  of  6  to  12: 
i.  e.  2. 

a.  Proportion  may  also  be  expressed  by  placing  the  sign  of  equal- 
ity between  the  two  ratios. 

Thus  :  2l  8  =  4;  16  is  a  proportion,  and  is  read,  the  ratio  of 
2  to  8  is  equal  to  the  ratio  of  4  to  16. 

What  is  a  Simple  ratio?  A  Compound  ratio?  Is  there  any  ditference  be- 
tween Compound  ratio,  and  any  other  ratio?  Why  then  is  it  used?  Of  what 
does  a  proportion  consist?  Give  an  example.  What  are  the  terms  of  the  coup^ 
lets  which  compose  the  proportion  called?  How  is  proportion  usually  ex- 
pressed? How  read,  when  expressed  in  this  manner?  How  may  it  otherwise 
oe  expressed? 


Art.    2.  RATIO    AND    PROPORTION.  225 

Obs.  4.  From  tliese  illustrations,  we  conclude  that  four  mem- 
bers are  in  proportion^  when  thejirst  has  tlie  same  ratio  to  tlie  second^ 
thai  the  third  has  to  the  fourth, 

Obs.  5.  The  learner  will  perceive  from  what  has  been  said,  that 
Ratio  and  Propoition  are  different : 

1st.  Because  a  ratio  consists  of  but  two  terms,  or  one  couplet ;  whilst 
a  pwportion  cannot  exist  without  four  terms,  or  two  couplets, 

2nd.  Because  one  ratio  may  be  greater  or  less  than  another;  thus, 
the  ratio  of  2  ;  8  is  greater  than  that  of  4  ;  12,  and  less  than  that 
of  3  :  15;  hui  in  a  proportion  the  ratios  must  be  equal,  and  conse- 
quently one  cannot  be  greater  or  less  than  another, 

Obs,  6.  Although  it  takes  four  ternu  to  fo:m  a  proportion,  still 
it  can  be  formed  with  three  Tmmbers,  as  one  of  the  numbers  may  be 
repeated. 

Thus  :3:6:  :6:12isa  proportion,  of  which  G  is  the  con- 
sequent of  the  first  couplet,  and  the  antecedent  of  the  second 
couplet. 

Obs,  7.  When  a  number  is  repeated,  as  in  this  example,  it  is 
called  a  mean  propioriional  between  the  other  two  numbers  ;  and  the 
last  term  is  called  a  third  proportional  to  the  other  two  numbers. 

Thus,  in  the  above  example,  6  is  a  mean  proportional  between  3 
and  12,  and  12  is  a  third  proportional  to  3  and  6. 

Obs.  8.  The  first  and  last  terms  of  a  proportion  are  called  the 
EXTREMES  ;  the  two  middle  terms  are  called  the  means. 

Thus:  in  the  proportion  15  :  3  ;  :  20  :  4,  15  and  4  are  the  ex- 
tremes, and  3  and  20  the  means. 

Obs.  9.  In  evej'y  proportion  the  product  of  the  extremes  is  equal  to 
the  product  cf  the  means  ;  otherwise,  they  are  not- proportional . 

Thus  .4:2:    :  6  :  3  is  proportional,  because  4  X  3  =  2  X  6. 
But  6:2;    :    12  :  3  is  not  proportional,   because  6  X  3=  18, 
and  12X2  =  24,  which  is  i  reater. 

How  read,  when  expressed  In  this  maimer?  When  are  four  numbers  in 
proportion?  Are  ratib  and  proportion  alike?  Why  not?  Of  how  many  dif- 
ferent numbers  can  proportion  consist?  How  can  this  be  done?  Wiiat  term 
in  each  couplet  is  the  number  repeated?  What  is  the  number  that  is  repeated 
called?  What  is  the  last  term  called?  T«  the  proportion  3  :  6  :  :  6  :  12,  which 
is  the  mean  proportional,  and  which  the  third  proportional?  Which  are 
the  extremes  in  a  proportion?  V\  hich  the  means?  In  the  proportion  15  :  3 
:  :  20  :  4,  which  are  the  extremes,  and  which  the  means?  Wiiat  principle  is 
true  with  regard  to  every  proportion?  If  this  is  not  the  case,  what  is  the  con- 
clusion? Is  4  :  2  :  :  6  J  3  proportional?  Why?  Is  6  ,•  2  :  ;  12  :  3  propor- 
tional? Why  not? 
*.  lU 


226  COMMON    ARITHMETIC.       '  Scct.  XU 

Jllus(railo7i  1st. — In  llie  proportion  4  ;  2  ;  I  6  I  3,  the  ratio  of 
4  :  2  is  ~ ;  also,  that  of  6  ;  3  is  ^  ;  hence,  the  ratios  of  the  two 
couplets  being  equal,  the  numbers  are  proportional,  according  to 
Obs.  1. 

2nd. — From  the  fact  that  proportion  is  an  equality  of  rat'os,  it  is 
evident  that  the  product,  either  of  the  extremes  or  means,  must 
contain,  as  factors,  the  antecedents  of  both  couplets  and  the  7'atio.  But 
the  consequent  of  each  couplet  is  equal  to  the  antecedent  of  this  coup' 
let  multiplied  by  the  ratio.  (Art.  1.  Obs.  6.)  Therefore,  the  pro - 
ducts  of  the  antecedent  of  each  couplet,  midtiplied  by  the  consequent  of 
tlw  other  couplet,  are  equal. 

Obs.  10,  As  in  Ratio,  so  in  Proportion,  the  two  terms  of  each 
couplet  must  be  of  the  same  denomination.   (Art.  1,  Obs.  8.) 

Rf.mark. — The  learner  will  observe  that  it  is  not  necessary  for  all  four  terms 
of  the  proportion  to  be  of  the  same  denomination,  but  only  those  of  each 
couplet.  For  $6  has  tiie  same  ratio  to  $3,  that  8  days  has  to  4  days  ;  that  is, 
the  ratio  of  each  is  2.  Therefore,  $6  :  $3  :  :  8  days  :  4  days:  is  a  correct  pro- 
ponion. 

Obs.  1 1 .  By  attentively  examining  Obs.  9,  we  notice  two  con- 
siderations : 

a.  1st.  Jjf  we  divide  the  product  of  the  means  by  either  extreme,  the 
quotient  will  be  the  other  extreme, 

b.  2nd.  If  we  divide  the  product  of  the  extremes  by  either  mean, 
the  quotient  will  be  the  other  mean. 

Thus,  in  the  proport" on,         5  :   10  :    :  6  :   12; 
If  we  have  given  the  ^ 

two     means,    and  v  12X5=60;  60-— 10=6,  one  mean, 
either  mean,  ^ 

Or, 12X5=60;  60-f-6=10,  the  other  mean. 

If  w^e  have  given  the  ^ 

two  extnmes,  and  >  10x6=60;  60-^-5=:  12  one  extreme, 
either  extreme,        )  i 

Or, 10X6=60;  60-M2=5,  the  other  extreme,  I 

Obs.  12.     Hence — If  any  three  terms  of  a  p)ro2Jortion  are  given^ 
the  other  may  be  found  ly  dividing  the  product  of  the  means  by  the    , 
given  extreme,  or  the  product  of  the  extremes  ly  the  given  mean.  | 

Explain  this  point  by  the  lilustrafioH  giv^u?  Proportion  is  an  equality  of  ra- 
tios ;  what  inference  is  deduced  froni  lliis?  To  what  is  the  couseqwentof  each 
couplet  equal?  What  do  we  qonclude  from  this  fact?  What  is  necessary  in 
boUi  ratio  and  proportion?  Is  it  necessary  for  all  four  of  the  terms  to  be  of 
tlie  same  denotniuation?  Which  then?  Give  an  example  illustrating  this 
point.  What  is  the  first  consideration  we  notice  from  examining  Obs.  9?  'I'he 
wecond?  If  we  haVe  given  three  terms*  of  a  proportion,  how  do  we  find  the 
'"ottrth' 


Art.    2.  RATIO    AND    PROPOIlTrON.  22'7 

Remark. — This  is  evident  from  the  fact,  that  in  all  such  cases  icp  hfive  the  pro- 
duct  of  two  factors,  and  one  factor  given  to  find  ihe  other,  which  is  always  found 
by  dividing  the  product  by  the  given  factor.     (Sect.  VI.  Art.  1.  Ob^.  16.) 

MENTAL   EXERCISES. 

1 .  If  the  extremes  are  4  and  6  anJ  one  mean  is  3,  what  is  the 
other  mean? 

2.  If  the  extremes  are  8  and  15,  and  one  mean  is  5,  what  is  the 
other  mean? 

3.  If  the  means  are  6  and  12,  and  one  extreme  is  8,  what  is  the 
other  extreme? 

4.  If  the  means  are  7  and  8,  and  one  extreme  is  4,  what  i^  the 
other  extre  i.e? 

5.  If  the  extremes  are  4  and  12,  and  one  mean  is  8,  what  is  the 
other  mean? 

6.  If  the  means  are  4  and  9,  and  one  extreme  is  3,  what  is  the 
other  extreme? 

7.  If  the  extremes  are  8  and  9,  and  one  mean  is  12,  what  is  the 
other  mean? 

8  If  the  means  are  4  and  10,  and  one  extreme  is  20,  what  is  the 
other  extreme? 

9.  Is  3  :  4  :  :  9  :  12  proportional?  Is  7  :  8  :  :  9  !  13  pro- 
portional?    Is  12   :  6  :    :  20   :   8  proportional? 

1  ?.  Is  21    :   7  :    :  24  :  8  proportional?     Is    10   ;    12  :    :   15  : 
18  proportional? 

11.  s  36  :  9  :  :  42  :  7  proportional?  Is  72  :  8  :  :  108  :  9 
proportional? 

12.  Is  40  :  50  :  :  70  :  90  proportional?  Is  40  :  60  :  :  60 
'.  90  proportional? 

Note. — The  learner  should  be  required  to  give  his  reasons  in  the  above  ex- 
amples, why  lliey  are  proportional  or  not,  as  the  case  may  be. 

Obs.   13.     Proportion  is  of  two  kinds — Simple  and  Compound. 
Case  1. — Simple  Proportion. 

Obs.  14,  Simple  Proportion  is  an  equality  between  two  simple 
ratios. 

Thus  :  6  :  2  :    :  9   ;  3  is  a  simple  proportion. 

Remark  1. — The  chief  use  of  Sirnple  Proportion  is  to  find  the  fourth  term  of 
a  proportion,  when  the  first  terms  are  given.  It  id  in  tairf  manner  that  it  is 
chiefly  applifd  to  practical  business. 

2.— Simple  Proportion  is  often  Ciiled  the  Single  Rule  of  Three,  b<-cause  three 

Show  why  this  is  correct.  How  is  Propor.ion  diviied?  Wliat  is  Simple 
Proportion?"  For  what  is  it  chiefly  used?  What  is  Simple  Proporlion  ofteu 
calM?    Why? 


228  COMMON    ARITHMETIC.  Sect.    XII 

terms  are  given  to  find  a  fourth.  Proportion,  however,  vv^e  consider  to  be  the 
most  appropriate  name,  because  we  have  one  couplet,  and  the  antecedent  of 
another  couplet  given,  to  find  such  a  consequent  of  the  latter  couplet,  that 
both  shall  have  the  same  ratio,  and  thus  constitute  a  proportion.  (Obs.  1.) 

Ex.  1.  If  4  yards  of  cloth  cost  $6,  Low  nmch  would  7  yds.  cost? 

Ans.  $10.50. 

Solution. — As  in  every  couplet  the  terms  must  be  of  the  same 
kind,  it  is  evident  that  the  numbers  4  and  7  form  the  terms  of  one 
couplet,  because  they  are  both  yards,  and  $6  is  one  term  of  another 
couplet,  and  the  third  term  of  the  proportion,  and  oS  this  term  is 
dollars,  the  fourth  term  must  also  be  dollars,  in  order  that  both  may 
be  of  the  same  kind.     Hence — 

Obs.  15.  In  every  proportion  the  third  term  must  be  of  the  same 
kind  as  the  answer, 

Wfe  now  wish  to  ascertain  which  of  the  other  numbers  is  the  first, 
and  which  the  second  term.  If  we  make  7  the  first  term,  and  4 
the  second,  the  proportion  stands,  7  yds.  :  4  yds.  I  :  $6  :  the 
fourth  term,  or  answer. 

Here  we  have  the  two  means  and  one  extreme  of  a  proportion 
given,  to  find  the  other  extreme.  Then  6X4=24;  24-^-7=2^, 
(Obs.  12,)  and  the  proportion,  when  completed,  is  7  yds.  ;  4  yds. 
:  :  f  6  :  $3f.  Thus,  we  make  7  yds.  cost  less  than  4  yds.,  which 
is  absurd.  Therefore,  we  will  make  4  the  first  term,  and  7  the 
second  term,  and  the  proportion  stands  thus  :  4  yds.  ;  7  yds.  ;  ; 
^6  :  the  fourth  term,  which  we  find,  as  above,  to  be  $10.50. — 
Hence — 

Obs.  16.  If  the  ansiver  ought  to  he  greater  than  the  third  termi 
make  the  greater  of  the  remaining  numbers  the  second  term,  and  the 
less  number  the  first  term  of  the  proportion. 

Remark. — This  is  evident  from  tlie  fact,  that  if  the  answer  should  be  greater 
than  the  third  term,  the  ratio  of  each  couplet  is  greater  than  unity;  (Art.  1.  Obs. 
15.)  and  therefore  each  consequent  is  greater  than  its  antecedent. 

The  following  form  is  usually  adopted  in  simple  proportion  : 

We  multiply  the  6  and  7  together,         yds.     yds.  $ 

and  divide  their  product  by  4,  which  4     :     7    I    :    6 

.gives  $f0.50,  as  the  fourth  term.  6 

4)42 

Hence—  Ans.   10^  =  $10.50. 

What  is  the  most  appropriate  name?    Why?    What  must  the  answer  be  in 


Art.    2.  RATIO    AND   PROPORTIOIf.  229 

When  three  terms  of  a  proportion  are  given,  to  find  the 
fourth  : 

Obs.  17.  Multiply  together  the  second  and  third  terms,  and  divide 
the  product  by  the  first  term.  (Obs.  12.) 

Note. — The  learner  must  bear  in  mind  that  we  cannot  multiply  two  con- 
crete numbers  together  ;  (Sect.  IV.  Art.  2.  Obs.  6.  Rem.  1.  and  Sect.  VI.  Art. 
1.  Obs  1 5.)  thus  in  the  above  example  we  multiply  6  and  7  together,  and  divide 
the  product  by  4,  because  the  ratio  of  $6  to  the  answer,  must  be  the  same  as  4 
yds.  to.  7  yds.     In  this  way  we  make  the  terms  all  abstract. 

Proof.— 10.50X4=42;  7X6=42.  (Obs.  9.)     Hence— 

To  prove  Simple  Proportion  : 

Obs.  18.  Midtiply  together  the  first  term  and  the  answer,  or  fourth 
term  ;  and  also  the  second  and  third  teimis  ;  if  the  two  products  are 
equals  the  work  is  correct,  (Obs.  9.) 

Questions   in  Simple   Proportion  can  be  worked   by    cancela- 
tion : 


The  above  example  is  worked  thus  : 

We  place  the  7  and  6  at  the  right  for  the 


7 


=  101 


same  reason  that  we  make  them  the  second 

and  third  terms  in  the  proportion,   and  we 

place  the  4  at  the  left  for  the  same  reason  that  we  make  it  the  first 

term.     Again,  the  second  and  third  terms  are  multipliers,  and  the 

first    term    is    a    divisor.  (Obs.     17.,    and   Sect.    Vil.    Art.     1. 

Obs.  3.) 

Analytic  Solution. — If  4  yds.  cost  ^6,  1   yard  will  cost  6-r-4= 
$1.50,  and  7  yds.  will  cost  $1.50X7=^10. "50. 


2.  If  8  lbs.  of  tea  cost  87,  what  will  5  lbs.  cost? 


Ans.  $4.37|. 


every  proportion?  Why?  If  the  answer  should  be  greater  than  the  third 
term,  which  of  the  remaining  numbers  do  we  make  the  first,  and  whish  the 
second  term  of  the  proportion?  Show  why  this  is  correct.  Wlien  three  terms 
of  a  proportion  are  given,  how  do  we  find  ihe  fourth?  Can  we  multiply 
concrete  numbers  too^ether?  How  do  we  avoid  this  in  Proportion?  How 
do  we  prove  Simple  Proportion?  Why  is  this  correct?  By  what  other 
method  can  questions  in  Proportion  be  performed?  How  are  the  numbers 
arranged  for  canceling?  Why  arranged  in  this  manner?  If  the  answer 
ought  to  be  less  than  \he  third  term,  which  of  the  remaining  numbers  do 
we  make  the  first,  and  which  the  second  terra  of  the  proportion?  Show 
why  this  is  correct. 


230  COMMON    AR[THMETI0.  Scct.    XII 

Operation. 
We  arrange   the  terms   as  in  the  first         lbs.     lbs.       $ 
example,  excepting  that   we  make   5  the  8    :  5  :    :  7 

second  term,  because  5  lbs.  will  evidently  7 

cost  less  than  8  lbs.     We  then  proceed  — 

according  to  Obs.  17.  8!35 

Hence—  '  Ans.  4f=.S4.37i. 

Obs.  19.  If  the  ansv>er  our/hi  to  be  less  than  the  third  term,  make 
the  least  of  the  remaining  numbers  the  second  term,  and  the  greater  num- 
hsr  the  first  term  of  the  proportion. 

Remark. — Tlii:^  is  evident  from  the  fact,  that  if  the  answer  sliould  be  less 
than  the  third  term,  the  ratio  ofeich  couplet  i^i  less  than  unity;  (Art.  1.  Obs.  14.) 
ani  therefore,  each  consequent  should  be  less  than  its  antecedent. 

Analytic  Solntlm.~li  8  lbs.  cost  $7,  a  lb.  will  cost  $7-^8= Sf, 
and  5  lbs.  will  cosi  $JX5=^V=^4'37i,  as  before. 

There  is  also  nnother  method  by  which  questions  in  simple  pro- 
portion may  be  performed,  which  may  be  thus  explained: 

A  proportion  must  consist  of  two  couplets.  (Obs.  5,  a  )  and  tlie^e 
couplets  must  both  have  the  same  ratio.  (Obs.  1.,  and  Obs.  5.  6.) 
Kow,  in  examples  in  simple  proportion  we  have  given  one  couplet* 
and  the  antecedent  of  another  couplet,  to  find  the  consequent  of  thiss 
latter  couplet.  (Obs.  4.  Rem.  1.)  But  the  consequent  of  any  coup- 
let is  equal  to  its  antecedent,  multiplied  by  the  ratio  ;  (Art.  \.  Oa  s 
6.)  therefore,  as  the  first  term  must  have  the  same  ratio  to  the 
second,  as  the  third  has  to  the  fnirth,  (Obs.  4.) 

Obs.  20.  The  fourth  term  of  a  proportion  may  be  found  by  mtd~ 
tiplying  the  third  term  by  the  ratio  of  the  first  to  the  second. 

Thus,  in  the  first  example,  we  have  the  proportioa  4  I  7  :  I  6 
:  the  fourth  term.  Here  the  ratio  of  4  :  7  is  J,  (Art.  1.  Obs.  2.) 
and  $6Xi='i^l0.50,  as  before. 

!n  the  second  example,  we  have  the  proportion  8:5:  \  1  \ 
the  fourth  term.  Here  the  ratio  of  8  :  5  is  |-,  (Art.  1.  Obs.  2.)  and 
$7X|-=^4.37i^,  as  before. 

3.  If  3  bu.  3  pks.  6  qts.  of  wheat  must  be  given  for  2  galls. 
3  qts.  I  pt.  of  wine,  how  much  wine  must  be  given  for  6  bu.  2  pks. 
of  wheat?  Ans.  4  galls.  2  qts.  1||-  pts. 

Of  how  many  couplets  must  a  proportion  consist?  What  relation  must  the 
ratios  of  these  couplets  havs  to  each  other?  What  have  we  given  in  exam- 
ples in  Simple  Proportion?  What  have  we  to  find?  To  what  is  the  conse- 
quent of  any  co  iplet  equal?  What  relation  have  the  four  terms  of  a  propor- 
tion to  each  oihjr?  By  what  other  method  then  can  the  foarih  terra  of  a  pro- 
portiou  be  found? 


^'A 


Art.    2.  RATIO    ANt>    PROPORTION.  231 


Operation, 

bu.     pks. 
3  ._  3  . 
4 

qts. 
.-  6 

bu.     pks.     galls,     qts. 

:    6  --  2  :  :  2  ..  3  _ 

4                      4 

pt. 
-    1 

1 5  pks. 
8 

26  pks.            1 1  qts. 
8                     2 

126  qts. 

208  qts.    :    :  23  pts. 
23 

624 
416 

126)4784(371^  pts.-:4  galls.  2  qts.  l|^  pts.     Ans. 
378 

1004 
882 

In  this  example,  it  is  first  necessary  to  reduce  the  first  and  second 
terms  to  one,  and  the  same  denomination,  and  also  to  reduce  the 
third  term  to  the  lowest  denomination  mentioned  in  it,  before  we  can 
proceed  further.  We  then  work  as  usual,  and  the  fourth  term  of 
course  is  pints,  which  is  the  denomination  to  which  the  third  term 
was  reduced.  (Obs.  10.)     Hence — 

Obs.  20.  If  the  first  and  second  feigns  contain  different  denomina- 
tions, reduce  both  to  one  and  the  same  denonmiation  ;  and  if  the  third 
term  is  a  compound  number,  reduce  it  to  the  lowest  denomination 
mentioned  in  it ;•  after  this,  proceed  according  ^o  Obs.  17  ;  the  result 
will  be  of  the  denomination  as  the  third  term,  when  reduced. 

Remark. — The  arrangement  of  the  terms,  as  in  the  preceding  examples,  is 
called  stating  the  question. 


Analytic  Solution. — If  126  qts.  pay  for  23  pts.,  1  qt.  will  pay  for 
_!__  of  23  pts.,  or  yV^  pts.  Again,  if  1  qt.  pays  for  y^^  pts.  208 
qts.  will  pay  for  208\imes  as  much,  and  -f^^  X  208  =  ^||^  pts.  = 
4  galls,  2  qts.  Ijj  pts.,  as  before. 

From  the  preceding  examples,  we  percelv^e  that  all  questions  ift 


If  the  first  and  second  terms  contain  different  denominations,  how  do  we 
proceed?  If  the  third  term  is  a  compound  number,  how  do  we  proceed?  Of 
what  denomination  is  the  re&ult? 


232  COMMON    ARITHMETIC.  Scct.    XII 

Simple  Proportion   are  solved  on  precisely   the    same    principles. 
Hence — > 

Obs.  22.  All  questions  in  Simjole  Proportion  can  he  solved  by 
Analysis. 

From  the  preceding  remarks  and  illustrations,  we  derive  the  fol- 
lowing 

GENERAL  RULE  FOR  SIMPLE  PROPORTION. 

I.  Mahe  that  number  the  third  term  which  is  of  the  same  kind  as 
the  answer  require  J.  (Obs.  15.) 

II.  Consider  from  the  nature  of  the  question  whether  the  answer 
should  he  greaJter  or  less  than  this  term. 

IIL  If  the  answer  ought  to  he  greater  than  the  third  term,  maJc^ 
the  greater  of  the  two  remaining  numbers  the  second  term,  and  th^ 
less  number  the  first  term.     (Obs.  lo) 

ly.  If  the  answer  ought  to  be  less  than  the  third  term,  make 
the  least  of  the  two  remaining  numbers  the  second  term,  and  the 
greater  number  the  first  term.     (Obs.  19.) 

Y .  Then  multiply  the  second  and  third  terms  together,  and  divide 
the  product  by  the  first  term ;  the  quotient  will  he  the  fourth  term, 
or  answer^  (Obs.  17)  which  is  always  of  the  same  denomination  as 
the  third  term.     (Obs.  21.) 

Proof,- — Multiply  the  first  and  fourth  terms  together,  and  also  the 
second  and  third  terms;  if  the  two  products  are  equals  the  work  is 
correct.  ^Obs.  18.) 

Remark  1. — If  either,  or  all  the  terms!,  are  compound  numbers,  the  first  and 
second  terms  must  be  reduced  to  the  same  denominations,  and  the  third  term 
must  be  reduced  to  the  lowest  denomination  mentioned  in  it,  before  the  multipli- 
cation or  division  can  be  performed    (Otis.  21.) 

2. —  If  the  \G.iruer  choose,  he  may  reduce  the  lower  denominations  to  fractions 
or  decimals  of  a  hisher,  instead  ol  reduciiijs:  the  higher  to  the  lower  denomina- 
tions. (Sect."  IX.  Art.  3.  Cases  2,  .3,  and  4.  Rules.) 

3. — We  can  often  shorten  the  operation  by  multiplying  the  third  term  by  the 
ratio  of  the  first  term  to  the  second.  (Obs.  20.)  When  the  ratio  is  an  integer, 
or  small  fraction,  this  method  is  preferable. 

4. — If  there  is  a  remainder  after  the  division  has  been  performed,  reduce  it 
to  the  next  lower  denomination,  and  divide  as  before. 

5. — This  rule  is  equally  applicable,  whether  the  numbers  are  Integral,  Frac- 
tional, or  Decimal. 

What  is  the  arrangement  of  the  terms  as  in  the  preceding  examples  called? 
How  can  all  questions  in  Simple  Proportion  be  solved?  What  is  the  rule  in 
Simple  Proportion?  What  \i  the  proof?  If  either  or  all  the  terms  are  com- 
pound numbers,  how  do  we  proceed?  By  what  other  method  can  we  proceed 
with  compound  numbers?  How  can  we  often  shorten  the  openition?  When 
is  this  method  preferable?  If  there  is  a  remainder  after  the  division  has  been 
performed,  how  do  we  proceed? 


Art.    2.  RATIO    AND   PROPORTION.  233 

HuLE — By  Cancelation  :  Consider  from  the  nature  of  the  ques- 
tion which  are  the  differefU  terms ;  then  after  placing  the  second  and 
third  terms  at  the  right,  and  the  frst  term  at  the  left,  cancel  as 
usual. 

Note. — The  above  Remarks,  (the  third  excepted,)  are  equally  applicable  to 
•Cancelatiou  or  Proportion. 

EXERCISES    FOR    THE    SLATE. 

1.  If  8  bushels  of  apples  cost  $4,  how  much  wiJ  14  bushels  cost? 

Ans,  §7. 

2.  If  9  bushels  of  peaches  cost  86.75,  how  much  will  12  bush- 
els cost?  Ans.  ^9, 

3.  If  14  yds.  joi  cloth  cost  $21,  bow  much  will  11  yds.  cost? 

Ans.  $16.50. 

4.  If  13  pencils  cost  $22.75,  how  much  will  9  pencils  cost? 

Ans.  $15,75. 
6.  If  $9  buy  12  bushels  of  apples,  how  many  bushels  will  815 
buy?  ^  Ans.  20. 

6.  If  824. 50  will  pay  for  14  saws,  how  many  saws  can  be  bought 
for  $14?  ■  Ans,  8. 

7.  If  6  men  c.an  do  a  piece  <of  w^ork  in  1 8  days,  in  what  time  can 
9  men  do  it?  Ans.   12  days. 

8.  If  12  men  can  do  a  piece  of  workki  21  days,  in  what  time  can 

7  men  do  it?  Ans.  36  days.;:,  o: 

9.  If  10  Ibsof  sugar  cost  $1.25,  liow  much  will  1  cwt.  cost?  .!  v 

Ans.  $12.50. 

10.  If  a  man  can  travel  75  miles  m  3  days,  how  far  will  he  tra- 
vel in  3  weeks?  Ans.  450  miles. 

Note. — The  learner  will  observe  that  he  travels  6  days  in  the  week. 

11.  If  2  cwt.  3  qrs.  15  lbs.  of  cofiPee  cost  $30.45,  how  much 
will  2  qrs.  22  lbs  cost?  Ans,  $7.56. 

12.  If  2  yds.  3  qrs.  of  cloth  cost  $5,50,  how  much  will  3  qrs.  2 
na.-cost?  Ans.  $1.75. 

13.  If  the  moon  moves  105°  24'  40"  in  8  days,  how  long  will  it 
take  it  to  perform  one  revolution,  or  360°? 

Ans,  27  d.  7  h.  43  m.4-. 

14.  If  a  pole  5  feet  high  east  -a  siiadow  3  ft.  -6  in.  in  length,  how 
high  must  a  pole  be  to  cast  a  Shadow  1 30  feet? 

Ans.    1854  feet. 

15.  If  7  yards  of  cloth  cost  $10.50,  how  much  will  15  bushels 
of  apples  cost,  if  -9  bushels  ef  apples  are  worth  3  yards  of  cloth? 

Ans.  $7.50. 

lathis  rule  applicable  to  either  inteo-ral,  fractioaal,  or  decimal  jiumbers? 
What*i8  the  rule  by  cancelation?  ~~--^ 


2S4  COMMON    ARITHMETIC.  Sect.    XIX 

16.  If  frctti  a  staff  tliat's  three  feet  long, 

A  shadow  five  is  made, 
What  is  a  steeple's  heighth  in  yards. 
That's  ninety  feet  in  shadt- T 

A.ns.   18  yards. 

17.  If  I  gi^e  8r5-for  the  us«  of  $200  a  certain  time,  how  much 
must  I  give  for  the  usy  ef  $700  for  the  same  time? 

Ans.  $52.50. 

18.  If  a  family  of  6  persons  consume  50  lbs.  of  flour  in  a  certain 
time,  how  much  will  they  consume  in  the  same  time,  if  6  persons 
more  are  added  to  the  same  family?  Ans.  91§  ibs. 

19.  If  a  man  can  perform  a  piece  of  work  in  16  days,  working  10 
hours  per  day,  how  long  will  it  take  him  to  perform  it  if  he  works 
12  hours  per  day?  .  Ans.   iSg  days. 

20.  If  84  bushels  of  oats  last  30  horses  7  days,  how  long  will  215 
bushels  last  them?  Ans.   17— days. 

Upon  carefully  examining  the  question,  it  will  \  e  perceived  that 
the  number  of  horses  is  not  one  of  the  terms  of  the  proportion . 

21.  If  the  interest  of  $600  for  12  months  is  $36,  what  is  the  in-^ 
te:est  of  the  same  SHim  for  9  months?  Ans,  $27.    -' 

Fractions. — 22.  If  f  of  a  yard  of  cloth  cost  J  of  a  dollar,  how 
much  will  —  of  a  yard  cost? 

I  yd..:  Hyd.  :    :  $1  :  ?H=«lll.   Ans. 

23.  If  y  of  a  barrel  of  flour  cost  $8,  what  will  \~  of  a  barrel 
cost?  ^  $4iJ. 

24.  If  J  of  a  pound  of  tea  cost  $~,  how  much  will  4  of  a  pound 
cost?  '  Ans.  $0.75. 

25.  If  ~  of  I  of  a  gallon  of  wine  cost  j^  of  a  dollar,  how  much 
would  ~  of  y  of  a  gallon  cost?  x\ns.  $0.92^. 

26.  If  $18y  buy  15yo  bushels  of  wheat,  how  many  bushels  cau. 
be  bought  for  $28i-^?  Ans.  22j|:.    | 

27.  If  4|  yds.  of  cloth  cost  ^b\.,  how  much  will  7j  qrs.  cost?    -■ 

Ans.  2yV- 

28.  If  215  biishels  of  oats  last  30  horses  17~  days,  how  long 
will  84  bushels  last  them?  Ans.  7  days. 

29.  If  I  of"  an  acre  of  land  cost  $\',  how  much  would  yj  of  an 
acre  cost?  Ans.  '1^8—. 

30.  If  a  railroad  cargoes  162.75  miles  in  7.75  hours,  how  far, 
at  this  rate,  Avould  it  travel  in  2.  5  days?  Ans.   Iv60  miles. 

31.  If  22f  )bs.  of  butter  cost  $4.74,  how  much  would  1 J  cwt. 
cost?  Ans,  $39.37^. 

32.  If  5.25  bushels  of  wheat  cost  $4 ..724-,  how  much  would 
8.37  bushels  cost?  ^        Ans.  $7,533. 

33.  If  12.45  pounds  of  tea  co^t  $9.3375,  how  much  would 
9 .  875  lbs.  cost?  Ans.  $7 .  4C f 


Art.  2.  feATIO    AND    PROPOUTIOJ?.  ^35 

34.  If  20. 125  acres  of  land  cost  $322,  Iioav  mncli  wdlild  15.5 
acres  cost?  Ans.  ^248, 

35.  If  12  horses  eat  62,75  bushels  of  oavl^  in  7  days,  how  much 
will  they  eat  in  15  days?  An^*.  134. 46"-4- bushels. 

36.  i^  Held  was  measui?e(?.  imd  found  to  be  r48  rod>  in  length  ; 
but  afterwards  the  line  with  wiiich  t  was  :i  easured  was  found  to  be 
but  31 1  feet  long  instead  of  33  feet  as  was  supposed.  Required — 
the  true  length  ©f  the  field?  Ans:   142i|  rds, 

Case  2. — Compound  Proportion: 

ObSw  ^.  A  COMPOUND'  PR0ifOiixiO5f  is-  an  equality  letween  a 
cvmpound.  cmd  a  simple  ratio. 

3  •  12  ) 
Thus:     4  .  ';o(  •   •  5  •  ^00  is  a  eompoimd  proportion,  and  is 

read^ — "  3"  into  ^  is  to'  >2  iato  20  as  5  is  to  100  ;  "  or,  '•  the  ratio  of 
3  x:4  :  12  X  20  is  equal  to  the  mtio-of  5  :  100." 

RkmarK  I. — The.  learnor  will  perceive  that  it  is  not  thft  ratio  of  3  :  12,  or 
that  of  4  :  20  alo.'ie,  which  is  equal  to  that  of  5  :  IGO  j  bu-t  t<  is  t\it.  ralvo  com- 
poundsd  sf  ^i^S€  which  is  equal  to  that  of  5  :  iOO.     Thus  : 

3X4  :  12X20  :  :  5  :  100  ;  because  3X4X100=12X20X5.— 
(Obs.  ^.) 

2. — Compownd  Proportion  is  cMsfly  used  tchen  more  than  one  staiementf  is  neC" 
essary  in  Simple  Proportion.  It  is  often  called  The  Double  Rule  of  Three,  but 
we  consider  the  term  Compound  Proportion  preferable,  as  it  ex|>resse8  what  it 
really  is. 

Ex.  1.  If  a  man  ride  246  mires  in  6  days,  riding  8  hours  per  day, 
how  far  caw  he  ride  in  14  days>  riding  12  hours  per  day? 

Ans.  861  miles. 

Solutton.-^-\\\  this  example  two  things  are  to  be  considered  : 

Tst.  The  number  of  days  he  rode.  From  this  we  deduce  this 
question:  If  a  man  rides  2'4'Q  miles  in  6  days,  how  far  will  he 
ride  in  \4  <^kfs.     And  the  proportion  and  result  stand  thus: 

6:14  days  :  :  246  miles  :  574  miles. 

2nd.  Th^  nnmher  of  hours  he  rode  jyer  day.  From  this  and 
the  last  py??.portion  we  obtain  this  question  :  If  a  man  rides  574 
miles  in  a  cei'lain  numJjcr  of  days,  riding  8  h&urs  per  day,  how  far 
will  he  ride  in  the  same  time,  riding  1 2  hours  per  day?  And  the 
proportion  and  result  stand  thus  : 

8  hrs.  :  12  hrs.  :  :  574  miles  .  861  miles. 


VVbat  is  Co4T! pound  Proporiion?  Plow  do  you  read  the  expression  given? 
Is  it  the  ratio  of  .^  :  12,  or  of  4'  :  20  alone  which  is  equal  to  that  of  5  ;  100? 
What  ratio  ili.'u  is  equal  lo  that  of  6  :  100?  Why?  When  is  Compound 
Proporiiou  chiefly  used?  What  is  it  often'  ealled?  What  ia  itiijare  appropri- 
ate uaine?     Why? 


.xii^ 


236  COMMON  arithmetic"'  "Sect 

All  questions  in  Compound  Proportion  involve  tlie  same  princi- 
ples.   Henae — 

Obs.  24.     AU  questions  in  Compound  Proportion  can  he  p&rfomied 
by  two  or  more  statements  in  Simple  Proportion. 

This  exan^le,  howerca*,  can  be  worked  by  one  statement,  thus : 

Operation. 

6  days    :     14  days    >    ..  246  miles  :  tlie  answer. 
8  hours  .     12  hours  ^ 

48  :  168  :  :  246  miles  :  the  answer. 

^46 

1008 
672 
336  ^ 


68)41328(861  miles.  Answer. 
:384 

292 
288 

48 
48 

00 

As  w€  desire  miles  for  our  answer,  we  make  246  miles  the  third 
term,  (Obs.  15.)  Then,  if  he  rides  246  miles  in  6  days,  he  will 
evidently  ride  farther  in  14  days;  therefore,  we  make  14  the  second 
term,  and  6  the  first  term.  (Obs.  16.)  Again,  if  he  rides  12  hours 
perd^y,  he  will  evidently  ride  fartliier  tlian  if  h^  rides  but  8;  tf;ere- 
fore  we  make  12  another  second  term,  and  8  the  first.  (Obs.  16.) 
But  we  perceive  this  to  be  a  compound  proportion,  (Obs.  23.)  and 
k  equivalent  to  the  simple  proportion  48  (6X8)  :  168(14x12) 
:  :  ii46  :  the  answer.  (Obs.  23.  Rem.  1.)  which  is  found  according 
to  Obs.  17.     Hence— 

To  perform  operations  by  one  statement  in  Compound  Pro- 
portion. 

Obs.  25.    Arrange  the  third  term,  and  each  couplet  as  in  Simple 

H6»(r  can  all  questions  in  Compound  Proportion  be  performed?  When  we 
make  but  one  stat-ement,  how  do  we  arrange  the  tejras? 


4jrt.  2.  RATIO    AMD   PROPORTION.  237 

Proportion^  (Obs.  15,  16,  and  17.)  Then  multiply  the  continued 
product  of  all  the  second  terms  by  the  third  tenn,  and  divide  the 
result  by  the  conUned  product  «f  aU  the  first  terms. 

The  abore  example  may  be  proved  tlins  :  6X8X861  =  14X1^ 
X246.  (Obs.  9.)     Hence — 

To  prove  Compound  Proportion  : 

Obs.  26.  Multiply  the  continued  products  of  the  first  terms  by 
the  fourth  term ;  also,  the  continued  product  of  the  second  terms  by 
the  ^ird  term ;  if  the  tiro  results  are  edike,  the  work  is  correct, 
(Obs.  9.) 

Remark.— la  all  questions  in  Compoond  Proportion,  there  is  one  number  of 
a  different  kiad  from  any  of  the  rest,  and  this  number  is  tdtoaifs  made  the  third 
term. 

The  fourth  term  of  the  above  proportion  may  also  be  found  thus: 
The  ratio  compounded  of  two  couplets  6:14  and  8:12  is  3J. 
(Art.  1.  Obs.  16.  b.)     Xow  the  third  term  (246)  must  have  tl- 
same  ratio  to  the  fourth  term,   or  answer,  (Obs.  23,  Rem,  1.)  and 
«46x3i=861.  (Obs.  20.) 

Hence — ^we  can  obtain  the  fourth  term  oi  a  Compound  Pro- 
portion : 

Obs.  27.  By  multiplying  the  third  <en»  by  the  ratio  compowuM 
of  the  first  and  second  terms. 

Analytic  Solution. — If  he  rides  246  miles  in  6  days,  he  will  ride 
J  of  246  miles,  or  41  miles,  in  1  day  ;  tlien  if  he  rides  but  8  hours 
per  day,  he  will  ride  i^  of  41  miles,  or  y  miles  per  hour ;  and  in 
12  hours  he  will  ride  12  times  as  far,  or  *  |^  miles  ;  and  in  1  •  days, 
of  12  hours  each,  he  will  ride  14  times  *|'  miles,  which  is  861 
miles,  as  before. 

All  questions  in  Compound  Proportion  are  solved  on  the  same 
principles.     Hence — 

Obs.  28.     All  questions  i»  Compound  Proportion  can  he  solved 

iy  Analysis. 

Operations  in  Compound  Proportion  can  frequently  be  shortened 
by  canceling  factors  common  to  the  first,  and  the  seccmd  or  third 
terms,  as  in  the  following 

How  do  we  next  proceed  ?  How  do  we  prove  the  operation?  What  occurs 
in  all  questions  in  Compound  Proportion?  What  term  is  this  number  made? 
By  what  other  method  can  the  fourth  term  of  a  compound  proportion  be 
found?  How  may  ail  questions  in  Compoand  Proportion  be  solved?  How 
caa  we  often  shorten  the  operation? 


2S8  r-oMMON  -AmTiiMETic.  Sect.  Xn 

Operation^ 

^  I  Xfi..7  0«r  first  krms   are  all  factors  of  some 

^..4-  -8     .l"^--^  number  v>^hich  is  n  divisor,  and.  our  second 

^;4(!>-.123       and    third   terms  are   all  factors   of  some 
123X7=BC1  immbvr  Avliich   is   a  dividend  ;  (Obs   25.) 
therefore  we  place  all  our  first  terms  at  the  left  of  the  line,  and  all 
the  other  numbers  at  the  right  of  the  line,  (Sect.  VIL  Art.  1.  Obs. 
3.)  and  cancel  as  usual. 

Rkmaiik  1. — We  have  prt^sented  lh.e?e  diffreiit  inetliods  of  obtaininf  tlye  re- 
sult, both  in  Simple  and  Compound  Proportion,  for  thf'  purpose  of  showing; 
the  relation  which  numbers  bp.ir  o  each  other,  and  also  to  show  how  the  same 
result  may  be  obtained  by  ditTfrent  processes,  thus  proving  clearly  the  correct- 
ness of  the  prjnc.p^esby  which  we  aperate.  And  tiii-s  we  consider  to  t>«  one  of 
the  principal  beauties  of  mathematics — Ihat  vre  are  obliged  to  take  no  principle, 
or  believe  no  theory,  unless  it  is  ct/p'ihle  of  dnnonstrntion,  or  proof.  And  one  in- 
dubitable proof  of  the  correctness  of  atiiematical  principles,  is,  that  we  al- 
ways obtain  the.  same  result  to  a  question  or  proidem,  no  matter  what  princi 
pies  we  apply,  or  what  process  ice  use,  promded  we  make  no  mistake  in  our 
reasoning,  or  mode  of  appllcalinn.  The  case  would  be  otherwise,  were  our 
principles  incorrect,  because  then  the  app^lication  of  diflerent  principles  to  the 
same  question,  w-ould  often  produce  different  results  ;  hut  as  this  never  hap- 
pens, we  conclude  that  they  are  absolutely  trnv. 

2. — Of  the  dilTerent  methods  of  obtaining  the  result  in  proportion,  that  by 
Analysis  is  decidedly  preferable,  as  it  is  belter  adapted  to  the  discipline  of  the 
mind,  and  strengthens  the  mental  faculties.  It  is  recommended  to  the  pupil  to 
solve  all  the  questions,  both  in  Si nple  and  Compound  Proportion,  by  Analysis, 
after  he  has  obtained  the  result  by  Proportion.  He  will  by  this  means  gain 
sufficient  mental  culture  to  amply  c-einpensate  him  for  his  trouble. 

From  the  foregoing  remarks  and  illusti^ations,  we  derive  the  fol- 
following  V 

GENERAL  RULE  FOR  COMPOUND  PROPORTIOxN.  ^ 

I.  3faJce  that  number  the  third  term  which  is  of  the  same  kind  as 
tli-e  ansiver.  (Obs.  15.) 

II.  Take  tux)  of  the  remaining  numhers  of  the  same  kind,  and 
arrange  them  in  the  same  manner  as  in  Simple  Proportion.  (Obs.  16 
and  17.)     Proceed  in  the  same  manner  with  all  the  other  numbers. 

IIL  Multiply  the  continued  product  of  the  second  terms  by  the 
third  term,  and  divide  tlie  result  by  the  continued  product  of  the  first 
terms.  (Obs.  25.) 

How  do  we  arrange  the  terms  far  canceling'?  Why  do  wearrange  them  i« 
this  manner?  What  do  these  different  methods  in  obtaining  the  result  in  Sim- 
ple and  Compound  Proportion  show?  What  does  this  prove?  Wliat  is  one  of 
the  principal  beauties  in  mathematics?  What  is  one  in:iubitable  proof  of  the 
correctness  of  mathematical  principles?  Would  this  be  so  were  our  princi- 
ples incorrect?  Why  not?  As  this  never  happen?,  what  do  we  conclude?  Of 
the  different  methods  of  obtaining  the  result  iu  Proportion,  which  is  preferabit! 
Why  80?     What  is  the  rule  in  Compound  Proportion? 


Art.    2.  RATIO    AND    PROPORTlOlf^.  23U 

Proof. — Multiply  the  fourth  term  by  the  continued  product  of  the 
first  terms  ;  also,  multiply  the  third  term  hy  the  continued  pjrodact  of 
the  second  terms  ;  and  if  the  two  residts  are  -equal,  ihe  vjorlc  is  cor- 
red.  (Obs.  26.) 

Note. — The  reitiafksundt?>ftib  rule  in  Simple  f^ropb'rtron  are  equally  appli- 
crfbiH  tD'Compounil  Pro()orlion,  except  the  third,  of  which  it  will  be  observed 
in  Compound  Proportion  that  we  mullipiy  the  third  term  by  the  ratio  com- 
pounded of  the  first  and  second  terms,  instead  of  th«  simple  ratio  of  either 
couplet.     (Obs.  27.) 

Rule — By  Cancelation. —  Consider  which  are  the  different  terms  : 
then  place  the  third  term  together  with  all  the  second  tenns  at  the  right, 
4ind  all  the  first  terms  at  the  left. 

EXERCISES    FOR    THE    SLATE. 

1.  If  15  men  earn  8225  in  12  days,  how  much  will  9  men  earn 
in  24  days,  at  the  same  rate?  Ans.  $270. 

2.  If  8  men  earn  $B4  in  7  days,  working  10  hours  per  day,  how 
long  will  it  take  12  men  to  earn  $300,  working  12  hours  per  day? 

Ans.   13|  days. 

3.  If  7  men  cut  49  acres  of  grass  in  4  days,  how  many  days 
will  it  take  11  men  to  cut  142  acres  of  grass?  xVns.  5~. 

4.  If  the  interest  on  8200  for  2  years  is  824,  what  is  the  interest 
on  8150  for  1  year  and  9  months?  Ans.   815.75. 

5.  If  12  students  spend  8100  in  3  weeks,  how  much  will  7  stu- 
dents spend  in  5  weeks?  /\ns.  897. 22§. 

6.  If  10  men  dig  a  trench  240  yards  long,  6  feet  wide,  and  5  feet 
deep,  in  8  days,  how  long  will  it  take  18  men  to  dig  a  trench  400 
yards  long,  8  feet  wide,  and  4  feet  deep?  Ans,  7Jy  days. 

7.  If  6  men,  working  12y\  hours  per  day,  dig  a  cellar  22^^  ft. 
long,  17y\  ft.  wide,  and  4|  ft.  deep,  in  2^  days,  how  long  will  it 
take  9  men  to  dig  a  cellar  45  ft,  long,  35|  ft.  wide,  and  5y\  ft.  deep, 
working  8j  hours  per  day?  Ans.   12  days. 

8.  If  80  men  dig  20  cellars,  each  45  ft.  long,  28  ft.  wide,  and  3 
ft.  deep,  in  10  days,  working  12  hours  per  day,  how  many  men  will 
it  require  to  dig  30  cellars,  each  24  ft.  long,  21  ft.  wide,  and  4^  ft. 
deep,  in  18  days,  working  but  10  hours  per  day,  supposing  the 
strength  of  the  men  in  the  former  case  to  be  Ij  that  of  those  in  the 
latter,  and  the  hardness  of  the  ground  in  the  latter  case  to  be  only 
\  of  that  in  the  former?  Ans.  56. 


What  is  Baid  respecting  the  remarks  under  the  rule  in  Simple  Proportion? 
What  must  be  observed  with  respect  to  remark  third?  Wi.at  is  the  rule  by 
c^neeiation? 


240  COMMON    ARHHMETiC.  Scct.    XII 

Article  3.     Partnership. 

Obs.  1.  Partnership  is  the  associatino;  toi^etlier  of  two  or  more 
persons  in  joint  trade,  with  a  mutual  agreement  to  share  the  respect- 
ive gains  or  losses  in  proportion  to  the  capital  each  one  has  invested, 
and  the  time  it  has  been  employed  ;  this  association  is  called  a  Com- 
pany, or  Firm,  and  each  individual  is  called  a  Partner. 

Obs.  2.  Stock  is  the  name  given  to  the  money,  or  value  of  the 
articles  employed  in  trade.     It  is  sometimes  called  Capital, 

The  gain  or  loss  divided  is  called  the  Dividend. 
There  are  two  cases  of  Partnership. 

Case  1. — This  case  is  used  when  no  consideration  is  made  with 
regard  to  time. 

Ex.  1.  A,  and  B.  enter  into  partnership,  A.  furnished  8300, 
and  B,  8500;  they  gained  8250.  What  was  each  man's  share 
of  the  gain? 

Solution.— K.  put  in  83C0,  and  B.  8500  ;  hence,  8300-4-8500== 
8800,  the  whole  stock  invested.  Now  it  is  evident  that  each  one's 
share,  of  the  gain  must  have  the  same  relat  on  to  the  whole  gain,  as 
each  one's  share  has  to  the  whole  stock.     Hence,  the  following 

Operation, 

Whole        A.'s     Whole      A.'s  share 
A.'s  stock.  stock.       stock,      gain*         of  gain. 

B.'s  stock.         Then,-8800  :  8300  :  :  8250  :  893.75. 

B.'s  stock.  B. 's* share  of  g'n 

8800  Whole  stock.     And,   8800  :  8500  ::  8250  :  8156.25. 

{^  93.75  A.'s  share  of  the  gain, 
p  J  8156.25  B.'s  share  of  the  gain. 


1^8250.00  Whole  gain. 

Solution  hij  Analysis. — As  A.  put  in  8300,  he  owns  ^J  J  =  J  of 
the  stock  ;  therefore  he  must  have  \  of  the  gain,  and  the  f  of  8250 
is  893.75,  A.'s  gain.  .. 

Also,  as  B.  put  in  8500,  he  owns  \%%  of  the  stock,  and  therefore 
must  have  |-  of  the  gain  ;  \  of  8250  is  8156.25,  B.'s  gain 

2.  Suppose,  in  the  above  case,  they  had  lost  8200  ;  what  would 
have  been  each  one's  share  of  the  loss? 

What  is  Partnership?  What  is  the  association  called?  What  is  each  indi- 
vidual called?  What  is  stock'?  What  is  it  sometimes  called?  What  is  the 
gain  or  loss  to  be  divided  called?  Of  hoW  many  cases  does  Partnership  con- 
sist?    When  is  Case  1  used? 


Art.    3.  RATIO    AND    PROPORTION.  241 

Operation^ 

^300 

$500     Then,  $800  :  $300  :  :  $200  :  $  75,  A.'s  loss* 

And     $800  :  $500  :  :  $200  :  $125.  B.'s  loss. 

$800 

f$  75,  A.'s  loss.  ■"-  -"''-^ 

-r>  I  $125,  B.'s  loss.  ,,ar 

Proof:      -^  '  <)i^ 


[  $200  whole  loss. 

Analytic  Solution. — A.  put  in  j  of  the  stock,  and  therefore  must 
sustain  J  of  the  loss  :  f  of  $200  is  $75,  A.'s  share  of  the  loss,  B. 
put  in  |-  of  the  stock,  and  therefore  must  sustain  f  of  the  loss  ;  f  of 
2OO  is  $125,  B.'s  share  of  the  loss. 

From  these  illustrations  we  derive  the  following 

RULE  FOR  PARTNERSHIP. 

As  the  whole  stock  is  to  each  man^s.  share  9/  the  stock,  so  is  the 
whole  gain  or  loss  to  each  man's  share  of  the  gain  or  loss. 

Proof. — Add  together  the  several  sJmres  of  the  gain  or  loss  i  if 
the  sum  is  equal  to  the  whole  gain  or  loss,  the  work  is  correct,  ri^ 

Remark. — This  case  is  often  called  Single  Fellowship  ;  but  as  a  partnership 
cannot  exist  without  at  least  two  individuals,  the  propriety  of  calling  it  single 
is  somewhat  doubtful.  , 

3.  A.,  B.,  a^id  G:  trade  together:  A.  puts  iii  $'60O,  B.  $800,  and 
.0.  $1000  ;  they  gain  $690.     What  is  each  one's  share. 

Ans.  A.'s,  $150  ;  B.'s,  $240  ;  and  C.'s,  $300. 

4.  A.,  B.,  and  C.  speculate  together.  A.  furnishes  $3000,  B. 
$6000,  and  C.  $4000 ;  they  lose  $3000.  How  much  was  each 
one's  loss?  Ans.  A's  $750  ;  B's  $1250  ;  and  C's  81000. 

5.  P.,  Y.,  and  R.  own  a  vessel  worth  $10000.  P's  share  is  worth 
$3500;  Y.'s  share  is  worth  $4250,  and  R.  owns  the  rest*  In  a 
certain  trip  they  gaii;  $1250.  How  much  is  each  one's  share  of 
the  gain?  *  '  ; 

Ans.  P.'s  $437.50;  Y.'s  $531.25;  and  R.'s  $281.2^;^ 

6.  U.  Y.  and  W.  own  a  store.  U,  owns  4,  Y.  owns  y\,  and  W. 
owns  the  rest;  they  lose  $480.  How  much  is  each  one!s  share  of 
the  loss?  .,  ./ 

Ans.  U.'s  $137.142|  ;  Y.  and  W.,  each  $171.4^84. 

7.  A.,  B.  and  C,  loaded  £^  ship  wth  flour,     A.  put  on  4C0  bar. 

What  is  the  method  of  proof.?  What  is  this  case  often  called?  Is  this  Utm 
approprififte?    Why  not? 

'  12  ^^: 


242  COMMON    ARITHMETIC.  ScCt.    XII 

rels,  B.  350,   and  C.  250;  in  a  gale  the  Captain  threw  overboard 
200  barrels.     What  Io6S  did  each  one  sustain? 

A.'s  loss  SObbls.;  B.'s  70  bbl.s;  and  C.'s  50  bbls. 

8.  Four  persons  trade  together.  A.  put  in  ^200  ;  B.  S4C0;  C. 
8350;  and  D,  $550  ;  they  gain  i600.     What  is  the  share  of  each? 

Ans.  A.'s,  $80;  B,'s,  $160;  C.'s  $140;  and  D.'s  ^220. 

Obs-  4,  Bankruptcy. — A  Bankrupt  is  a  person  who  is  unib!e 
to  pay  his  debts. 

Questions  in  Bankruptcy  are  usually  performed  by  the  rule  of 
Partnership. 

9 .  A  bankrupt  owes  $4000 ;  his  property  is  worth  but  $3000. 
How  much  can  he  pay  on  the  dollar?  Ans.  $0.75. 

10.  A  man's  property  is  worth  $5000.  He  owes  A.  $1000,  B. 
$3500,  C.  $850,  D.  $1500,  and  E.  $2150.    How  much  can  he  pay 

each  in  proportion  to  their  debts? 

^  Ans.  A.  $555.555J;  B.  $1944. 444i;  C.  $472.222j;  D, 
I       $833.3331;  and  E.  $1194. 444i. 

11.  A  man  died,  leaving  property  worth  $10000.  He  owed  F. 
$2500;  G.  $3050;  H.  $4500;  and  I,  $2450.  How  much  can  each 
creditor  receive,  and  how  much  can  the  estate  pay  on  the  dollar? 

J  Ans.  F.  receives  $2000;  G.  $2440;  H.  $3600;  and  I.  $1960. 
\      And  the  estate  pays  $0. 80  cents  on  the  dollar. 

Obs.  5.  General  Average. — When  the  master  of  a  vessel,  in 
consequence  of  a  storm,  or  other  casualties  at  sea,  is  obliged  to 
throw  overboard  a  part  of  the  cargo,  in  order  to  save  the  ship  and 
crew,  the  law  requires  that  the  loss  shall  be  sustained  by  tht^  own- 
evs  of  the  vessel  and  cargo,  ia  proportion  to  the  value  of  each  indi- 
vidual's property  at  stake.  The  property  sacrificed  is  called  the 
Jettison. 

The  process  of  finding  each  one's  loss  in  such  cases  is  called 
General  Average.     The  rule  is  the  same  as  in  Partnership. 

12.  A  vessel   being  in  distress,    the  Captain  threw   ovei board 
goods  to  the  value  of  $15000      The  cargo  was  owned  by  thr  e  per- 
sons, L.,  M.  and  N.    L.  owned  $20000;  M.  S35000;  and  N.  $25000: ' 
the  vessel  was  worth  $20000.     How  much  was  each  man's  loss? 
J  Ans.  L.'s  $3000;  M.'s  $5250;  N.'s  $3750;  and  the  owner  of 
I     the  vessel  $3000. 

What  is  a  bankrupt?  How  are  questions  in  bankruptcy  usually  performed? 
When  the  master  of  a  vessel  is  obliged  to  throw  overboard  a  f»art  of  his  cargo, 
in  order  to  save  his  ship  or  crew,  how  is  the  loss  sustained?  What  is  ihe  pro- 
o«M  of  fiading  each  one's  loss  ia  such  cases  called?    What  is  the  rule? 


rt.    3  PARTNERSHIP.  243 

1 3.  In  consequence  of  a  storm  at  sea,  a  vessel  sustained  the  fol- 
lowing loss : 

"'.  For  a  part  of  her  cargo  thrown  overboard,  valued  at     $4850 

'     For  repairing  damages  of  the  ship,  &c.,  1 450 

Other  expenses, 300 

The  loss  was  sustained  by   a  general  average,  the  value   of  the 
ship  and  cargo  being  as  follows  : 

Value  of  the  vessel, . 825000. 

Part  uf  the  cargo  owned  by  Wra.  Trader,  <fe  Co,  valued  at      48000, 

Part  "         "         "      Thos  Dealer, &  Co., "         loOOO, 

Part  **         ''         "      James  Lovevain, '*         27000. 

Part  "         **         ''       Captain  of  the  vessel,  ___ ''         10000. 

What  share  of  the  loss  must  each  sustain? 

C  Ans.   The  owner  of  the  vessel  -S 11 20;  Tra  'er  &  Co.  $21 50. 40; 
<  Dealer  &  Co.  $672;  James  Lovegain  #1209.60;  Captain 

(  of  the  vessel  $448, 

A  number  can  also  be  divided  into  proportional  parts  by  the  rale 
of  Partnership. 

14.  Divide  the  number  576  into  three  parts  that  shall  be  to  each 
other  as  the  numbers  7,  8,  and  9. 

7_|_8+9  =  24;  24  :  7  :  :  576  :  168,   &c. 

Ans.   168,  192,  and  216. 

15.  Divide  the  number  1728  into  6  parts,  that  will  be  to  each 
other  as  the  numbers  3,  4,  5,  6,  8,  and  10. 

Ans.   144,  192,  240,  288,  384,  and  480. 

Case  2. — Obs.  6.     7'his  case  is  used  when  time  is  considered. 

Ex.  1.  A.  and  B.  hire  a  pasture  for  810.  A.  put  in  two  cows 
4  months,  and  B.  put  in  4  cows  3  months.  How  much  ought  each 
to  pay? 

Solution. — A.  put  in  2  cows  4  months;  this  is  the  same  as  8  cows 
1  month.  B.  put  in  4  cows  3  months  ;  this  is  the  same  as  12  cows 
1  month.     Hence,  they  must  };ay  in  the  proportion  of  8  to  12. 

Operation.  :,    ^ 

2X4=    8.  -^^ 

4X3=12      Then  20:    8  ::  810  :  84  A. 's  share.  T 

—      And  20  :  12  :  :  810  :  ^6  B,'s  share.  ^  ow 

20  1^ 


Proof. 


4  A.'s  share. 
6  B.'s  share. 


inf>o 


810  Whole  cost. 
When  is  Case  2  used? 


244  COMMON    AEITHMKTIC.  Sect.    XIII 

Analytic  Solution. — A.  has  in  the  same  as  8  cows  1  month,  and 
B.  has  in  the  same  as  12  cows  1  month;  and  both  have  in  the  same 
as  8  -{-  12  ==  20  cows  1  month.  Then  A.  must  pay  ^V  =  |>  and  B. 
1|  =  I  of  the  cost.  I  of  $10  is  .f  4,  A.'s  share,  and  |  of  ^10  is 
%Q,  B.'s  share,  as  before.     Hence — 

To  solve  questions  in  Partnership,  when  time  is  considered  :      .r, 

Obs.  7,  Multiply  each  marl's  stock  by  the  time  he  continues  it  in 
trade,  and  use  the  product  for  his  share. 

2.  A.,  B,  and  C.  trade  together,  A.  put  in  $400  for  6  months; 
B.  put  in  $500  for  4  months;  and  C.  put  in  $350  for  8  months  ; 
they  gained  $300.     How  much  was  each  one's  share  of  the  gain? 

Ans.  A.'s  $100;  B.'s  $83.3331-;  and  C.'s  $116.666|. 

3.  Suppose  in  the  last  example  they  had  lost  $200.  How  much 
would  have  been  each  one's  share  of  the  loss? 

Ans,  A.'s  $66. 666|;  B.'s  $53.555|;  C.'s  $77.777j. 

4.  W.,  X.,  Y.  and  Z.  trade  together  for  one  year.  W.  put  in 
$500  for  6  months,  and  then  put  in  $700  more  ;  X.  put  in  $1200 
for  4  months,  nnd  then  took  out  $500;  Y.  put  in  $800  for  5  months, 
and  then  took  out  $200,  but  put  it  back  again  at  the  end  of  9 
months,  with  $400  more;  Z.  put  in  $700  for  7  months,  and  then 
put  in  $800  more  ;  but  he  took  out  the  whole  of  his  stock  at  the 
end  of  1 1  months  ;  at  the  end  of  the  year  they  have  gained  $2490, 
How  much  is  each  one's  share?  :  *,,, 

Ans.  W.'s  $612;  X.'s  $624;  Y.'s  $600;  and  Z.'s  $654.' 


SECTION  XIII. 

PERCENTAGE. 

Article  1.     Definitions,  Mental  Exercises,  &c. 

Obs.  1.  The  terms  percentage  mid.  per  cent  are  derived  from  two 
Latin  words  jt?<?r  and  centum,  which  signify  hy  the  hundred. 

Therefore,  when  we  speak  of  a  certain  per  cent  of  any  number, 
we  evidently  mean  such  a  hundredth  part  of  that  number. 

Thus,  5  per  cent,  of  any  number  is  yj^  of  that  number;  12  per 
cent  of  any  number  is  j--^  ^^  ^^^t  number,  &c. 

How  do  we  solve  questions  in  Partnership  wiien  time  is  considered?  From 
what  are  the  terms  per  cent  and  percentage  derived?  What  do  these  words  sig- 
nify? What  do  we  mean  when  we  speak  of  a  certain  per  cent  of  any 
number?  w" -^    '•-''    Vwiii. 


Art.    1.  PERCENTAGE.  245. 

MENTAL   EXERCISES. 

1.  What  is  2  percent  of  100? 

2  per  cent  of  100  isyl^  of  100.  Now  ^^^  of  100  is  1,  and 
■j-lo-  of  100  is  1  X  2  =  2.  Ans.  2. 

2.  How  much  is  4  per  cent  of  100  cents?  Ans.  4  cents. 
4  percent  is  yjo-     yj-o  of  100  cents  is  4  cents. 

3.  A  man  borrowed  ^100,  paying  6  per  cent  for  the  use  of  it. 
How  much  did  he  pay?  Ans.  $6. 

6  per  cent  is  yj^,     y|-^  of  $100  is  $6. 

4.  A  man  having  100  sheep,  lost  10  percent  of  them  by  disease. 
How^  many  did  he  lose? 

10  per  cent  is  -,Vo  •     yVo  of  100  is  10.     Hence— 

Obs.  2.  Per  cent  implies  so  many  units  for  each  100  units,  so 
many  cents  for  each  100  cents,  so  many  dollars  for  each  100  dollars, 
so  many  articles  for  each  100  articles,  &c.:  and  the  per  centage  of  any 
number  is  as  many  times  the  per  cent  as  there  are  hundreds  in  this 
number. 

5.  How  much  is  6  per  cent  of  300?.  Ans,  6X3=18. 

6.  How  much  is  3  per  cent  of  200?  400;  700;  500;  800;  900 
290? 

7.  How  much  is  8  percent  of  100?  200;  500;  700;  900;  400 
800;   1200? 

8-  How  much  is  10  per  cent  of  200?  1200f  1500;  900;  700 
1000;  2000? 

9.  A  man  having  found  $6,  the  owner  g  ive  him  4  per  cent  for 
finding  it.     How  much  did  he  receive?  Ans.  24  cents. 

Solution.— E.e  received  4  cents  for  each  100  cents  ;  in  $6  are  600 
cents.     Therefore  he  received  4  X  6  =  24  cents. 

10.  A  merchant  sold  a  coat  lor  $12;  his  gain  was  10  per  cent,  of 
what  he  sold  it  for.     What  was  his  gain? 

11.  Wliat  is  10  per  cent  of  ^5'1  of  $8;  of  ^10;  of  ^11;  of  $6; 
of  f  9;  of  ^20? 

12.  A  constable  collected  $8,  and  received  5  per  cent  for  his 
services.     How  much  did  he  receive? 

13.  How  much  is  5  per  cent  of  ^2?  Of  ^4?  Of  810?  Of  7? 
Of  5?    Of  $12?     Of  $61 

What  is  5  percent  of  any  number?  12  percent?  6  per  cent?  3perceiit? 
9  pt^rcenl?  1  per  cent?  VVh;it  is  2  pt-r  cent  of  100?  6  per  cent  of  100 
cents?  9  per  cent  of  100  cents?  8  per  cent  of  100?  2  per  cent  of  $100? 
12  per  gent  of  $100?  What  does  per  cent  imply?  Wliat  is  tiie  percentage 
of  any  liumber? 


216  COMMON    ARITHME'nC.  Scct.  XIII 

14.  A  man  had  200  sheep  ;  in  two  years  they  increased  50 
per  cent.     How  many  had  they  increased? 

15.  An  auctioneer  sold  #1200  worth  of  gco  Is,  and  received  4  per 
cent  for  selling.     How  n.uch  did  he  receive? 

16.  What  is  4  per  cent  of  ^200?  Of  8800?  Of  ^600]  Of 
81000?     Of  $300? 

17.  A  man  borrowed  8460,  and  gave  6  per  cent  for  the  use  of 
it?     How  much  did  it  cost  himl 

18.  How  much  is  6  per  cent  of  8200?  Of  8800?  Of  8300? 
Of  8500?     Of  1200? 

19.  A  man  bought  a  farm  for  81000,  and  sold  it  so  as  to  gain  12 
per  cent?     How  much  did  he  gain? 

20.  What  is  12  per  cent  of  "^800?  Of  81200?  Of  8700?  Of 
81000?  Of  8500?  Of  8900?  Of  8300?  Of  600?  Of  8200? 
Of  81500?     Of  82000? 

Obs.  3,  Since  1  per  cent  signifies  y^^,  2  per  cent  i  Jo,  <£t.,  it  is 
evident  thai  it  may  be  expressed  decimaUy.     Thus  : 

1  per  cent  may  be  written .01. 

2  per  cent         "         "  .02. 

3  percent         "         ."    .  ,._ .03. 

4  per  cent         '*  *•  .04. 

7  per  cent         **  "  .07. 

9  per  cent         "  "         .09. 

a.  When  the  given  per  cent  is  10  or  more,  it  is  merely  ivritten 
with  the  decimal  point  before  it.     Thus  : 

10  per  cent   is  written .10. 

12  per  cent  "  .12,        " 

15  per  cent  **  .15. 

25  per  cent  **  -25.        .-= 

50  per  cent  *'  • -50. 

75  per  cent  **  .75. 

99  per  cent  '/ .99. 

6,  When  the  given  per  cent  is  100  or  more,  it  is  evidently  a 
mixed  number,  and  must  be  written  accordingly.     Thus  : 

100  per  cent  is  written 1.00,  or  1. 

104  per  cent      *'  1.04, 

125  per  cent       "  1.25. 

274  per  cent       '*  2.74. 

How  ctii  Huy  percetif.  !>;^  expn^ss' 1?  S'i>  v  whv  t!iis  \a  corri^ft?  When  the 
givei)  per  cent  is  10  or  more,  how  in  it  vvritltMi?  When  the  given  p«r  ceut  is 
IOJ  or  more,  what  kind  of  au  expression  i«  it?     How  written? 


Art.     1.  PERCENTAGE.  247 

c.     When  the  given  per  cent  is  less   than  1,  it  is  either  written 
ractionally ,  or  t  occupies  three  or  more  decimal  places.     Thus : 

*,  per  cent,  that  is  2  of  1  per  cent  is  written  _.  .00^^,  or  .005. 
i  per  cent,  that  is  J^  of  1  per  cent  is  written  __  .00^:,  or  .0025. 
I  per  cent,  that  is  |  of  1  per  cent  is  written  _.  .00 J,  or  .0075. 
^\  per  cent,  is  written  __  .02^,  or    .025&C 

21.  Write  5  percent;  6  per  cent;  8  per  cent ;  11  per  cent;  14 
per  cent ;  1 8  per  cent ;  22  per  cent ;  39  per  cent ;  50  per  cent ; 
an  I  18  per  cent. 

22.  Write  120  per  cent ;  150  per  cent  ,•  130  per  cent ;  \  per  cent 
and  Yu  V*^^  cent  i .  decimals. 

23.  Write  |  per  cent ;  yV  per  cent ;  y\  per  cent ;  and  -^^  per 
cent  in  decimals. 

per  cent;  4^  per  cent ;  6^  per  cent ;  8 J  per  cent ; 

.  in  decimals. 

I  per  cent ;  326 ^^  per  cent ;  and  51 2|  per  cent  in 


cent  in  decimals. 

24.   Write  2'^  per  cent;  A\  per  cent ;  6^^   per  cent ;  8 J 
ami  25V  per  cent  in  decimals. 

*,;  25.  Write   82|  per  cent;  326;i:  percent;  and  512|  j 
decimals. 


EXERCISES    FOR    THE    SLATE. 

1.  A  constable  collected  $600,  for  which  he  was  to  receive  4  per 
cent  for  collecting.     How  much  did  he  receive?  | 

Operation. 
4  per  cent  is  expressed    .04*     Therefo-e,   if  we  $600 

multiply  $600  by  .04,  the  product  wUl  be  what  the  .04 

constable  must  receive.  (Obs.  2.)     Hence —  

Ans.  624.00 
To  find  the  percentage  of  any  number  : 

Obs.  4,  Multiply  the  given  number  by  the  given  per  cent  ex- 
pressed decimally,  and  point  off  in  the  product  as  in  multiplication 
of  decimals.  (Sect.  VIII.  Art*  11.  Rule.) 

2.  What  is  6  per  cent  of  $80?  Of  $100?  Of  $150?  Of  $300? 
Of  $1000? 

3.  What  is  5  per  cent  of  $30.26?  Ans.  $1,513. 

4.  What  is  10  per  cent  of  $75.84?  Ans.  $7,584. 

5.  Wliat  is  15  per  cent  of  $125.75?  Ans.  $18,861.  _ 

6.  A  broker  exchanged  $1000,  tor  which,  he. was  to  reoeive  i-  per 
cent.     How  mnch  did  he  receive?  .-rr  Ans.  $5. 

7.  What  is  \  per  cen.  of  $2000?  Of  $Mo?  Of  $4000?  Of 
$3000?  Ans.  $5;  $2;  .'JfJlO;  $7.50. 

8.  What  is  i  per  cent  of  -450. 80?  Ans.  $1 .  127. 

When  the  given  ptr  cent  is  less  than  1,  how  is  it  written?  How  do  we  find 
the  per'ceuttLfire  "<"  ••*.>'  r?wi»>her? 


248  COMMON     ARITHMETIC.  Scct.    XIII 

9.  What  is  1  per  cent  of  $400?  Of  $800?  Of  $1200?  Of 
$1600?  Ans.  $.60;  $1;  $1.60;  $1.87i^. 

10.  What  is  I  per  cent  of  $1000?  Ans.  $6.26. 

11.  What  is  i  per  cent  of  $600?  Ans,  $2. 

Remark. — When  the  given  per  cent  is  a  fraction,  we  may  first  find  the  per 
centage  at  Iper  cent,  and  then  take  fractional  parts  of  this  for  the  given  per  cent. 
This  method  is  generally  preferable. 

12.  What  is  \  per  cent  of  $1200?     Of  $1600?     Of  $1800? 

Ans.  $2;  $2.60;  $3. 

13.  What  is  A\  per  cent  of  $200?  Ans.  $9. 

14.  What  is  3^  per  cent  of  $60?  Of  $80?    Of  $100?    Of  $200? 

Ans.  $1.92;  $2.66;  $3.20;  $6.40. 

15.  An  auctioneer  sold  $1000  worth  of  goods,  for  which  he  was 
to  receive  12  per  cent*    How  much  did  he  receive?     Ans.  $120. 

Note. — The  learner  will  bear  in  mind  that  he  received  12  per  cent  on 
what  he  sold,  and  not  on  what  wn.'i  paid  over ,  as  is  often  supposed.  In  the  lat- 
ter c&se  he  would  only  rec.^iveyf  2'  instead  of  yo  0"  of  the  value  of  the  goods 
sold.     The  same  remark  is  applicable  to  moneys  collected  by  constables,  &c^ 

16.  A  merchant  bought  $16000  worth  of  goods,  and  sold  them 
so  as  to  gain  16  per  cent.     How  much  did  he  gain? 

Ans.  $2260. 

17.  \  clock  dealer  shipped  800  clocks  for  New  Orleans,  but  on 
the  passage  12  per  cent  of  them  were  washed  overboard.  How 
many  were  lost?  Ans.  96. 

18.  Two  men  engage  in  trade  with  $1600  apiece.  One  gains 
!I6  p(  r  cent,  and  the  other  gained  20  per  cent.  How  much  did  one 
gain  R  ore  than  the  other?  Ans.  $60. 

!9.  A  gentlemen  deposited  $1200  in  a  bank,  and  afterwards 
drew  out  19|-  per  cent  of  it.  How  much  did  he  take  out,  and  how 
luucli  had  he  left? 

Ans.  He  took  out  $234,  and  had  left  $966. 
'  '20,  What  is  the  difference  between  18  per  cent  of  $300,  and  24 
per  cent  of  $400?  Ans.  $42. 

21.  What  is  the  sum  of  14  per  cent  of  $200,  and  9  per  cent  of 
$700?  Ans.  $91. 

22.  What  is  112  per  cent  of  200;  800;  1200;  1600;  18241 
4276;  6000? 

Ans.  in  order.     224;  896;   1344;   1680;  2042.88;  4789. 12;  5600. 

23.  What  is  103  percent  of  400?  Ans.  412. 

24.  What  is  260  per  cent  of  100?  400;  600;  900;  1400;  1846; 
8120? 

Ans.  in  order.     260;   1000;   1500;  2260;  3600;  4615;  20300. 

When  the  given  per  cent  is  a  fmction,  how  do  we  proceed?  Is  the  percent- 
age received  by  constables,  auctioneers,  &c.,  calculated  on  the  value  of  the 
property  they  sell  or  collect,  or  on  what  they  pay  over?     Why  bo? 


Art.  2.  COMMISSION  and  insurance.  249 

25.  What  is6i^  per  cent  of  $240?  6273.80;  ii?320;  8480.40; 
^512.96? 

Ans.  in  order.     $15;  $17. IH;  $20;  $30.02^;  $32. 06. 

26.  What  is  Si  per  cent  of  $24^J?  $345;  $462.36;  $512.34; 
$680.04? 

Ans.  in  order.     $20;  $28.75;  $38.53;  $42. 69^;  $56.72. 

27.  What  is  37^  per  cent  of  $250?  $300;  $412.25;  $560.59; 
$618,901 

^Ans.  in  order.     $93.75;  $112.50;  $154,593+;  $210. 18|; 
I  $232. 08  J. 

28*  Which  is  the  most — 8  per  cent  of  $500,  or  6  per  cent  of 
$700?     How  much  is  the  difference? 

Ans.  6  per  cent  of  $700  is  the  greatest  by  $2. 

29.  What  is  the  diflPerence  between  4^  per  cent  and  6^  per  cent  of 
^1500?  Ans.  ^32.50. 

30.  What  is  i  per  cent  of  $200?  $240.75;  $300;  $312.25 
$473.95? 

Ans.  in  order.  $0.40;  $0.48+;  $0.60;  $0,624+;  ,C:0.947+ 

31.  What  is  -jV  per  cent  of  $120?  $160;  $190;  $210;  $235^ 
$375.50? 

^Ans.  inorder.       $0.84;    $1.12;    $1.33;    $1.47;     $1,645 
}  $2,628+. 

32.  What  is/o  per  cent  of  $400?  $800;  $1000;  $1200;  $1500 
$2400? 

Ans,  inorder.     $1.80;  $3.60;  $4.  50;  $5.40;  $6.75;  $10.80 
33.   What  isj\  per  cent  of  $48?  $72;  $84;  $120;  $288;  $324 
$1728. 

^Ans.  inorder.  $0.20;  $0.30;  $0.35;  $0.50;  $1.20;  $1.35 
}  $7.20. 

34.  A  drover  having  300  cattle,  sold  40  per  cent  of  them.  How 
many  did  he  sell?  Ans.   120. 

35.  Two  men  had  each  $1500.  One  gained  20  per  cent  of  his, 
and  the  other  spent  20  per  cent  of  his.  How  m.uch  then  had  one 
more  than  the  other?  Ans.  $600. 

Ohs.  5.  Percentage  is  applied  to  various  calculations  in  practical 
business-  The  most  important  of  these  are  Commission,  Insurcmce, 
Profit  and  Loss,  Stocks,  Brokage,  Duties,   Taxes,  and  Interest. 

Article  2.     Commission   and  Insurance. 

Obs,  1.  Commission. —  Commission  is  the  per  cent,  or  sum  charged 
by  an  agent  for  transacting  business  for  his  employer.  It  is  generally 
applied  to  buying  aad  selling  goods. 

To  what  is  Percentage  applied?     Which  are  tiie  most  important  of  th 
Whajt'is  Commission?     To  what  is  it  generally  applied? 

12a 


250  COMMON    ARITHMETIC.  ScCt.    Xfll 

Remark  1. — The  person  who  buys  orsells goods  foranotheris  called  a  Com' 
mission  Merchant,  Correspondent,  or  Factor.  '"^ 

2. — As  ccinmission  is  reckoned  at  so  much  per  cent  on  the  money  employed 
the  rule  is  the  aame  as  in  percentage.     (Art.  1.  Obs.  4.) 

EXERCISES    FOR    I  HE    SLATE. 

1.  A,  sold  5^200  worth  of  goods  for  B.,  at  6  per  cent  commission 
How  much  did  he  receive?  Ans.  812. 

2.  A  man  sold  100  bushels  of  wheat  at  $1.25  per  bushel,  on 
commission,  for  which  he  received  5  per  cent.  How  much  was  his 
commission?  Ans.  $6.25. 

3.  How  much  is  the  commission  for  selling  ^^500  worth  of  goods, 
at  8  per  cent?  Ans.  $40. 

8.  How  much  is  the  commision  for  buying  $1200  worth  of 
goods,  at  4  per  cent?  Ans.  $48.^ 

5.  B.  sold  on  commission  for  C.  $427.80  worth  of  goods,  at  8|- 
per  cent;  $512.40  worth  of  goods  at  10^^  per  cent;  and  $350.50 
worth  of  goods  at  12|  per  cent.  How  much  did  his  commission 
amount  to?  Ans.  $132,914. 

6.  D.  sold  on  commission  for  E.  120  yards  of  cloth,  at  $4.50  per 
yard,  for  which  he  was  to  receive  4i-  per  cent.  How  much  was  his 
commission,  and  how  much  must  he  return  to  E.? 

i  Ans.  The  commission  was  $23.40,  and  the  sum  returned  was 
I  $516.60. 

7.  A  Southern  merchant  consigned  to  a  merchant  in  Boston  the 
following  art  cles,  to  be  sold  on  commision,  at  7|  per  cent :  4200 
lbs.  of  cotton,  at  7^  cts  per  pound  ;  7600  lbs.  of  sugar,  at  8j-  cts. 
per  pound;  19  bags  of  c  .'fFee,  each  containing  90  lbs.,  at  7  cents  per 
pound;  2100  lbs.  of  rice,  at  4 J  cents  per  pound;  and  25  barrels  of 
molasses,  at  $15  a  barrel.  How  much  was  the  commission,  and 
what  sum  must  be  returned  to  the  owner? 

i  Ans.  The  commission  was  $1 16.77,  and  tlie  sum  returned  was 
I  $1419.68. 

INSUllANCB. 

Obs.  2.  By  INSURA^X'E  is  meant  security  froiR  loss  or  dmiage 
occasioned  hy  fires,  storms ,  shipwrecks,  <{:c.  ^ 

Obs.  3.  This  security  is  usually  effected  with  imdividuais,  or  in- 
surance cojnjjanies,  by  the  payment  of  a  stipulated  sum,  which  is 
generally  a  certain  per  cent  of  the  value  of  the  property  insured. 


vVhfst  is  the  person  who  buys  or  sella  goods  for  others  ca!ic*d?  How  is  con*^ 
mission  reckon«"l?  What  tht^n  ifi  thf^  rul-'?  What  i«  rnt-ant  by  i!i.*urnnce? 
How  is  the  security  effVcleci?  When  an  insuriince  is  effected  by  u  rontract 
with  individuals,  what  is  it  termed?  Wluit  is  Mariue  Insurance?  How  ef- 
fectwi? 


Art.  *2.  roMMfssioN  and  insurance.  251 

Remark  1. — When  insurance  is  effected  by  a  contract  with  individuals,  it  is 
termed  out  door  insurance.  Insurance  at  sea  is  called  Marine  Insurance.  It  is 
usually  effected  for  a  certain  voyage. 

2. — Tlje  insurers,  whether  individnals  or  an  incorporated  company,  are  of- 
ten cali^rd  under  writers. 

Obs.  4.  The  sum  paid  for  the  insurance  is  called  the  Pre- 
miuxm;  and  it  is  paid  when  the  insurance  is  effected. 

The  written  instrument  thai  binds  the  contracting  parties  is  called 
the  Policy. 

The  amount  of  pecuniary  responsibility  taken  by  the  insurers  is 
called  the  Risk.  It  is  sometimes  equal  to  the  whole  of  the  estimated 
value  of  the  property  insured,  and  sometimes  it  is  equal  to  only  a 
part  of  the  estimated  value. 

Case    1 . —  When  the  premium  is  a  certain  per  ecnt  of  the  value  of 
the  property  insured. 

Remark. — Tlie  vry  nature  of  this  case  shows  that  the  rule  is  the  same  as 
in  Percentage.  We  must  add  the  cost  of  the  policy  to  the  percentage,  how- 
ever, to  find  the  actual  cost  of  the  insurance. 

EXERCISES    FOR    THE    SLATE. 

1.  How  much  must  be  paid  for  insuring  a  house  valued  at  ^1200, 
at  4|-per  cent,  and  the  cost  of  the  policy  being  ^1.        Ans.  ^55. 

2.  How  much  must  be  paid  lor  insurance  on  a  store  valued  at 
$4000,  and  on  the  goods  valued  at  '$13vX)0,  at  3|  per  cent;  the  pol- 
icy costing  ^  1 .  501  Ans.  ^647 .  50. 

3.  How  much  must  be  paid  for  insuring  a  steamboat  valued  at 
$10000,  the  policy  costing^!,  and  .hepr  mium  being  5  per  cent  on 
J  of  the  value  of  the  boat,  that  being  the  amount  of  risk  taken  by 
the  company?  Ans.  $401, 

4.  How  much  must  be  paid  f  r  insuring  a  ship  valued  at  $20000, 
and  the  cargo  valued  at  $50000,  the  premium  on  the  ship  being  Q~ 
per  cent  on  \  of  its  estimated  value,  and  the  premium  on  the  cargo 
beirg  4i'^o  per  cent  on  J  of  its  estimated  value,  and  two  policies  be- 
ing required,  each  costing  $1?  Ans  $2858,25. 

Remark. —  In  mvlual  inaurance  companies  each  one  gives  a  premium  note  of 
so  much  per  cent  on  the  property  which  he  wishes  to  insure,  the  ritte  being  de- 
termined by  the  amouui  of  Risk.  A  certain  per  cent  is  paid  down  on  these 
notes  for  immediate  use,  and  any  losses  that  occur  more  than  this  per  cent  are 


What  are  the  insurers  often  called?  What  is  the  Premium?  When  paid? 
What  is  the  Policy?  The  Risk?  To  what  is  the  Risk  equal?  What  is  the 
rule  when  the  premium  is  a  certain  per  cent  of  the  Vdlue  of  the  property  m» 
aurekUV 


252  COMMON    ARITHMETLC.  ScCt.    XIII 

averaf^ed  on  the  preinium  notes,  and  thus  each  one  helps  to  supply  the  deficiency 
in  proportion  to  the  amount  of  property  he  has  insured.  The  aaiouut  of  the 
premium  notes  forms  Use  capital  of  the  company. 

5.  How  much  must  be  paid  for  insuring  a  store  valued  $15420, 
the  premium  note  being  8  per  cent,  and  the  assessments  2,  2|-,  4, 
3^,  and  l^per  ceitt? 

Opercdion. 
815420 
.08 


Si 233. 60     premium  note. 

.  131  sum  of  assessments. 


61680 
370080 
123360 


Ans.      $166.5360 
By  taking  8  per  cent  of  '*Si  15420,  we  find  our  premium  note  to  be 
$1233.60.     The   sum  of  the  several  assessments   is   13^*     Then 
13^  per  cent  of  C' 1233. 60  gives  8166  ,  536  as  our  answer. 

6.  How  much  must  be  paid  for  insuring  $8540  worth  of  pro- 
perty, at  7  per  cent,  the  assessments  being -J,  1,  \\,  J,  2,  and  \  per 
cent?  Ans.  $38,109. 

7.  Ho\v  much  must  be  paid  for  insuring  812500  worth  of  pro- 
perty, at  i2|-  pel  cent,  the  assessments  being  2,  2^,  3^-,  4,  |,  3 J, 
and  I  per  cent?  Ans.  ^263.02. 

8.  A  gentleman  paid  n^50  annually  for  insurance  on  hispropert}^ 
at  4  per  cent.  How  much  was  the  vahie  of  the  property  covered  by 
the  policy? 

Solution. — As  the  value  of  the  property   insured  multiplied  by 
the  rate  per  cent  of  insurance,  gives  the  premJum,  it  is  plain  that 
t  e  premium  divided  by  the  per  cent  will  give  the  value  of  the  pro-,^ 
perty   insured.       Therefore,    i50  -^  .04  =  .'^1250.  -^^ 

Ans.  $1250. 

Proof. — ^1259  X  -^4  =  $53  the  premium.     Hence — 

When  we  have  given  the  premium  and  the  rate  per  cent  of  instil? 
ranee,  to  find  the  value  of  the  property  insured:  '':^^ 

Obs.   5.     Divide  the  loremiimi  b'j  the  rate  per  cent  of  insurance 

What  is  the  method  pursued  by  mutual  insurance  co  npanies?  How  do  we 
find  the  value  of  the  property  insured,  when  we  have  given  the  premium  and 
he  rate  per  cent  of  insurance?  '  ^m* 


Art.  2.  COMMISSION  and  insurance.  253 

expressed  decimally,  and  point  off  in  ilie  quotient  as  in  Diviaion  of 
decimals.     (Sect.  Vlil.  Art.  12.  Rule.) 

9.  If  I  pay  $270  premium,  at  3  per  cent,  Tvhat  is  tlio  value  of 
tLe  property  insured?  Ans.  ^9000. 

10.  A  merchant  paid  $1000  premium  on  Lis  property,  at  5  per 
cent.     What  was  the  value  of  his  property?  Ans.  {$20000. 

11.  A  gentleman  paid  ^1500  premium,  at  3  per  cent,  on  a  ship 
and  cargo.     How  much  were  they  worth?  Ans.  «f  60000. 

12.  A  man  paid  850  premium  on  his  property,  valued  at  $1260. 
What  was  the  rate  per  cent  of  insurance?  Ans.  4. 

Solution. — As  the  value  of  the  property,  multiplied  by  the  rate 
per  cent  of  insurance,  gives  the  premium,  it  is  evident  that  the  pre- 
mium divided  by  the  value  of  the  property  will  give  the  per  cent. 

Therefore,  50 -f- 1250  =  ylf^  =  .04.     Hence— 

When  the  premium  and  value  of  the  property  are  given,  to  fmd 
the  per  cent  of  insu.ance  : 

Gbs.  6.  Mche  the  premium  the  numerator,  and  the  value  of  the 
property  the  denominator  of  a  common  fraction  ;  then  reduce  it  to  a 
decimal,     (8ect.  VHI.  Art.  9.  Obs.  4,) 

Note. — The  pupil  will  perceive  that  these  three  cas^s  of  insurance  recipro- 
cally prove  each  otl.er.J 

13.  A  gentleman  paid  8400  premium  on  his  property,  which  was 
valued  at  i  16000,     What  was  the  per  cent  of  insurance? 

Ans,  2|. 

14.  A  ship  captain  paid  $-1250  premium  on  his  vessel,  valued  at 
$25000.     What  was  the  per  cent  of  insurance?  Ans,  5. 

15.  A  merchant  paid  $1800  premium  on  his  storo  and  goods, 
valued  at  $60000.     What  w^as  the  per  cent  of  insurance? 

Ans.  3. 

Qase  2. —  When  a  person  wishes  to  insure  such  a  sum  that  in  case 
of  a  loss,  he  may  receive  both  the  value  of  his  property,  and  the 
price  for  insuring  : 

Ex.  1.  A  gentleman  wishes  to  insure  his  property,  w^orth  $1140, 
at  5  per  cent,  so  that  if  it  is  destroyed,  his  policy  will  cover  both 
the  premium  and  value  of  the  property  insured.  What  sum  must 
his  policy  cover? 

/Solution. — The  rate  of  insurance  is  5  per  cent ;  therefore  on  $100 

How  do  we  find  the  rate  per  cent  of  insurance,  when  we  have  given  the 
premium  and  the  value  of  the  property  insured?  What  relation  have  these 
ihrertcases  of  insurance  to  each  other? 


SSi  COMMON    ARITHMETIC.  Scct    XIIT 

he  receives  biit  $"95,  as  he  pays  ^5  for  insuring.  Therefore,  he  re- 
ce'ves  y~V  of  the  property  he  insures.  Then  ^1140  is  j~  of  the 
sum  necessary  to  be  insured,  and  '*$1140  is  --^-^  of  ^1200. 

Ans.  $1200. 
Proof. — $1200  X  '05  =  $60,  the  premium  he  would  pay, 
$1200  —  UO  =  '^1 140,  the  value  of  his  property.     Hence— 

To  solve  all  such  questions  : 

Obs.  7.  Multiply  the  sum  necessary  to  he  insured  by  100,  and 
divide  the  'product  by  100  diminished  by  the  rate  per  cent  of  iri- 
suring. 

2.  What  sum  must  I  have  insured,  at  B  per  cent,  to  cover 
$2760?  Ans.  $3000. 

3»  A  merchant  sent  a  cargo  to  Liverpool,  valued  at  '$12600. 
What  sum  must  he  have  insured  at  8  per  cent,  th  t  in  case  of  a 
wreck  he  mav  sustain  no  loss  by  the  operation? 

Ans.  $13586. 95i|. 
4.  What  sum  must  be  insured  at  10  per  cent,  in  order  to  cover 
both  the  premium,  and  $10800  worth  of  property? 

Ans.  $12000. 

ApwTicle  3.     Profit  and  Loss. 

Obs.  1.  Profit  and  Loss  in  trade  signify  the  sum  g<xined  or 
lost  in  common  business  transactions.  They  are  calculated  at  a  cer- 
tain per  cent  on  the  purchase  price,  or  sum  paid  for  the  articles  un- 
der consideration. 

MENTAL    EXERCISES. 

1.  A  man  bought  a  sack  of  wheat  for  $5,  and  sold  it  at  a  profit 
of  10  per  cent»     For  how  much  did  he  sell  it? 

Solution. — As  he  sold  it  at  10  per  cent  profit,  he  sold  it  for  $5, 
together  with  10  per  cent  of  $5.  Now  10  per  cent  of  $5  is  $0. 50, 
(Art.  1.  Obs.  4.)  and  $5  4- $0.50  =  $5.  50.  Ans.  $5.50. 

2.  Suppose  in  the  last  case  he  had  sold  the  wheat  at  10  per  cent 
loss.     How  much  would  he  have  received  for  it? 

Solutio7i.—10  per  cent  of  $5  is  $0. 50.  Then  if  he  sold  it  at  10 
per  cent  loss  he  sold  for  50  cents  less  than  it  cost  him,  and  $5  — 
$0.50  =  $4.50.  Ans.  $4.50. 

3.  Bought  a  stack  of  hay  for  $12,  and  wish  to  sell  it  at  8  per 
cent  profit.     For  how  much  must  I  sell  it? 

What  is  the  rule,  when  a  person  wishes  to  insure  such  a  sum  that  in  case  of 
aloes  he  may  receive  both  the  value  of  his  property,  and  the  price  paid  for  in- 
suring?   What  do  Profit  and  L«sb  signify?    How  calculat«'d? 


Art.    3.  PROFIT    AND    LOSS  255 

4.  C.  boii^lit  apiece  of  cloth  for  8 10  ;  but  it  being  (l;miiip;f<J,  he 
is  wilHnij  to  sell  it;  at  4  per  cent  loss.  For  how  much  must  he 
sell  it? 

6.  A  farmer  bought  som  •  hogs  for  ^11,  and  Avishes  to  sell  thera 
at  12  percent  profit.     For  how  much  must  he  sell  them? 

6.  A  man  bought  a  coat  for  ^8,  and  was  otiered  12jper  cent  for 
his  bargain.     How  much  was  he  offered  for  it? 

7.  A  merchant  bought  a  barrel  of  flour  for  85,  and  sold  it  at  5 
per  cent  loss.     For  how  much  did  he  sell  it? 

8.  A  man  bought  some  tea  for  $6,  and  sold  it  at  20  per  cent  pro- 
fit.    For  how  much  did  he  sell  it? 

9.  A  man  bought  some  paper  for  Si 2,  and  sold  it  at  8^-  per  cent 
profit.     For  how  much  did  he  sell  it? 

10.  A  boy  bought  a  cap  for  83,  and  sold  it  at  3^  per  cent  loss. 
For  how  much  did  he  sell  it? 

11.  A  man  bough-:  some  land  for  ^50,  and  sold  it  at  6  per  cent 
loss.     For  how  much  did  he  sell  it? 

12.  A  drover  having  400  sheep,  lost  9  per  cent  of  them  by  dis- 
ease.    How  many  had  he  left.^ 

EXERCISES    FOR   THE    SLATE. 

Case   1, — To  find  at  what  2^ rice  to  sell  an  article  in  order  to  gain 
or  lose  a  certain  per  ceiU,  the  purchase  price  being  given, 

1.  A  man  bought  a  wagon  for  860,  and  sold  it  at  5  per  cent  pro- 
fit.    For  how  much  did  he  sell  it,  and  how  much  did  he  gain] 
Operati(y)\. 

%  60  By  multiplying  the  purchase  price 

.05  ^7  the  per  cent,  we  find  his  profit  to 

be  i3.     Then  if  he  gained  $3,  he 

$  3.00  gained.  must  have  sold  it  for  ^60  -|-  ^3  = 

60.00  863. 


Ans.  863.00  selling  price. 

2.  A  man  bought  a  farm  for  $1200,  and  sold  it  at  8  per  cent  loss- 
How  much  did  he  lose,  and  for  what  price  Jdid  he  sell  it? 

Operation. 
%  1 200  8 1 200  purchase  price. 

.08  96  lost. 


Ans.  $96.00  $1104  selling  price. 

8  per  cent  of  $1200  is  $96.  Then  as  he  sold  it  at  8  per  cent 
loss,  he  evidently  sold  it  for  $96  less  than  he  gave  for  it;  and  $1200 
—  $96  =  $1 104.     Hence— 


256  COMxMON    ARITHMETIC.  ScCt.   XHl 

To  find  at  what  price  to  sell  aa  article  in  order  to  gain  or  lose  a 
certain  per  cent,  the  purchase  price  being  given  : 

Obs.  2.  First,  multiply  the  inircliase  price  hy  the  per  cent ;  this 
will  give  the  gain  or  loss.  Then  add  the  gain  to,  or  subtract  the  loss 
froin,  the  purchase  price,  as  the  case  may  require  ;  the  resuU  in  either 
case  will  be  the  selling  price. 

Remark.—As  100  per  cent  is  1,  or  unity,  it  is  evident  that  any  per  cent  pro- 
fit is  equal  to  100  increased  by  this  per  cent,  and  any  percent  loss  is  equal  to 
100  diminished  by  this  per  cent.     Hence,  the  selUng  price  may  be  found  thus  : 

Obs,  3.  Add  the  per  cent  gained  to,  or  subtract  the  per  cent  lost 
from  100,  and  multiply  the  purchase  price  by  the  sum  or  remainder, 
as  the  case  may  be,  and  point  off  two  decimals  ;  the  result  in  either 
case  will  be  the  selling  price. 

Note. — The  learner  will  bear  in  mind  that  it  is  only  the  selling  price  that  we 
obtain  by  this  method,  and  not  the  sum  gained  or  lost. 

The  operations  of  Ex.  1  and  2,  by  this  method,[  may  be  seen,  as 
follows  :  .  "  "  -  -^ 

Ex.  1.  Ex.  2, 

-^^60  81200 

100+5=105  100—8=         92 


% 


300  2400 

60  10800 


S63 .  00  selling  price.  f  1 1 04 .  00  seUing  price. 

3,  A  merchant  bought  tea  for  SO.  50  per  pound,  and  sold  it  at  20 
per  cent  profit.     For  how  much  did  he  sell  it  per  pound? 

Ans.  $0.60. 

4.  A  merchant  bought  a  lot  of  goods  for  S450  ;  but  getting  them 
damaged,  he  sold  them  at  10  per  cent  loss.  How  much  did  he  lose, 
and  for  how  much  did  he  sell  them? 

Ans.  He  lost  <^45,  and  sold  them  for  8405. 
4.  A  di  overbought  a  lot  of  sheep  for  $500,  and  sold  them  at  25 
per  cent  profit.     How  much  did  he  gain,  and  for  how  much  did  h© 
sell  them?  Ans.- He  gained  $i'25,  and  sold  them  for  $625. 

6.  Bought  a  span  of  horses  for  $250,  and  sold  them  at  15  per 
cent  profit.     Required — the  profit  and  the  selling  price. 

Ans.  Profit,  $37.50;  selling  price,  $287.50. 

How  do  we  find  at  what  price  to  sell  an  article  in  order  to  gain  or  lose  a 
certain  per  cent,  the  purchase  price  being  given?  By  what  other  rule  can  the 
selling  price  be  found?     Demonstrate  this  rule.  i.i^,<ilI'S^ —• 


Art.  3.  PROFIT    AND   LOSS.  257 

7.  A  man  having  a  flock  of  200  sheep,  lost  40  per  cent  of  them 
by  disease.     How  many  did  he  lose,  and  how  many  had  he  left? 

Ans.  He  lost  80,  and  had  left  1 20. 

8.  A  merchant  bought  cloth  for  $6  per  yard,  but  getting  it  dam- 
aged he  is  willing  to  sell  it  at  10  per  cent  loss.  For  how  much 
must  he  sell  it  per  yard?  Ans.  ^5.40. 

9.  Suppose  he  w^ished  to  gain  5  per  cent  on  the  above  cloth.  For 
how  much  must  he  sell  it  per  yard?  Ans.  $6 .  30. 

10.  If  he  w^ould  gain  15  per  cent,  at  what  price  per  yard  must 
he  sell  it?  Ans.  ^6.90. 

11.  A  merchant  bought  500  bales  of  cotton,  each  bale  containing 
225  lbs.,  at  9 1  cents  per  pound  ;  but  the  price  declining,  he  sold  it 
at  4i  per  cent  loss.  How  much  did  he  lose,  and  at  what  price  did 
he  sell  it?  Ans.  Lost  S475.31^;  sold  it  for  S10193.43. 

13.  A  dealer  in  grain  bought  48  loads  of  wheat,  each  containing 
44  bushels,  at  $1 .  12-J-  per  bushel;  SO  loads  of  rye,  each  containing 
52  bushels,  at  $0.62|  per  bushel;  115  sacks  of  oats,  each  contain- 
ing 15  bushels,  at  $0.20  per  bushel;  and  230  loads  of  corn,  each 
containing  35  bushels,  at  $0.28  per  bushel:  he  sold  the  entire 
quantity  so  as  to  gain  16|-  per  cent  on  the  cost.  Required — his 
gain,  and  the  price  at  which  he  sold. 

Ans.  Gain,  $991. 66|;  selling  price,  $6941. 66|. 

Case  2. — Two  numbers  being  given,  one  of  which  is  regarded  as  a 
certain  per  cent  of  the  other,  to  find  the  rate  per  cent. 

1.  What  per  cent  of  $20  is  $5? 

Solution. — From  the  nature  of  the  question,  $5  is  some  per  cent 
of  $20  ;  that  is,  $20  multiplied  by  some  number,  will  produce  $5. 
Therefore,  we  have  $5  as  the  product,  and  $20  as  one  factor,  and 
the  other  factor  must  be  $5-^$20  =  ^V  =  -25.  (Sect.  VI.  Art. 
1.  Obs.  16.) 

Proof.— $20  X  -25  =  $5.     Hence— 

To  find  the  per  cent  in  such  cases : 

Obs.  4.  Make  the  number  ivhich  is  regarded  as  the  percentage  the 
numerator,  and  the  number  of  which  it  is  so  much  per  cent  the  de- 
nominator of  a  common  fraction,  ami  reduce  this  to  a  decimal.  (Secf. 
VIII.  Art.  9.  Obs.  4.) 

2.  What  per  cent  of  $40  is  $1 1?  Ans.   .275  =  27|. 

3.  What  per  cent  of  $90  is  $30?  Ans.  33i. 

4.  What  per  cent  of  $80  is  $40?  Ans.  50. 

When  two  numbers  are  given,  one  of  which  is  rpgarded  as  a  certain  per 
cent  of  the  other,  how  do  we  find  the  per  cent? 


258  COMMON   AHITHMETIC.  Scct.    XIII 

6.  What  per  cent  of  120  is  6?  Ans.  5. 

6.  What  per  cent  of  1  is  2?  Ans.  200. 

7.  What  per  cent  of  60  is  73?  Ans.   125. 
8  What  per  cent  of  $5  is  5  cent;  ]  Ans.   1. 

9.  What  per  cent  of  $80  is  1^120?  Ans;   150. 

10.  What  per  cent  of  $40  is  5  cents?  Ans.    .00125  =  j.     ' 

11.  What  per  cent  of  $P0  is  G  cents?  Ans.  yV-     • 

12.  What  per  cent  of  $450  is  !5^0.90?  Ans.  ^. 

The  principles  of  this  case  are  applied  to  the  transactions  of 
business  as  follows : 

13.  A  merchant  boitirht  some  cloth  at  $5  per  yard,  and  sold  it 
■  for  ^6. 25  per  yard.     What  per  cent  did  he  gain?  , 

Operation.  ■ 

As  he    bought  for   $5,  and  sold   for  ^6. 25  selling  price. 

S6.25,   he   evidently  gained    $6.25 —  5.00  purchase  price. 

^5  =  ^1.25  on  each  yard.     The  ques-  

tion  then   is,    What  per    cent   of  $5   is  $1.25  gain  per  yard. 

^1.25?     By    Obs.    4  we  find    this    to  .  i||  =  25  jDer'cent. 
be  25. 

'      Proof.— 85  X  .25=  $1.25;  85  + $1 .25  =  •*$'6.25.   Hence— 

When  we  have  given  both  the  purchase  and  the  selling  price,  to 
find  il:e  gain  or  loss  per  cent  on  ihe  cost  or  purchase  pi  ice  : 

Obs.  5.  I'^ind  the  gain  or  loss  on  the  article  hy  subiracilng  one  price 
from,  the  other ;  make  this  the  mmierator,  and  the  cost  or  purchase 
jnice  the  de7iominator  of  a  conmwn  fraction  ;  then  reduce  this  to  ^ 
decimal,  (Sect.  Vlll.  Art.  9.  Obs.  4.)  ■' 

Ki;"MARK. — Tho  learner  will  perctive  that  tiie  ;;or(;i;ii<aot>  ]«.  oalciilatt-d  on  the 
co-t,  01  sum  paid,  for  the  article,  and  nut  on  ihe  aciH dp- price,  as  is  often  sup- 
po.si'd. 

11.  B<jiig!!t  a  wagon  lor  860,  and 'sold  it  for  870.  Required — 
the  per  cent  gained.  Ans.   16f. 

15.  A  man  bought  a  horse  for  8t^0,  and  sold  him  for  •'ti'75.  Re- 
quired— the  gain  per  cent.  Ans.  25. 

16.  A  tailor  bought  some  cloth  for  '.r.  120,  and  sold  it  for  tl.S5. 
Required — the  gain  per  cent.  Ans.   12-V. 

17.  A  man  bought  a  carriage  for  1^200,  and  sold  it  for  $180. 
Require::— tile  loss  pe   cent.  Ans.   10. 


When   we  l)ave  given  botli  the  purchase  a  )!  s-Ming  pric^,  how  (io  we  fin  i 
the  gain  or  Iosh  per  cent?     1«  the  perceiilage  cuiculaletl  ou  the  buying  or  sell- 


Art.  3.      *  pfe(5nT  and  Lost^.  259 

18.  A  grocer  bjjglit  butter  for  18  cts.  per  pound,  and  sold  it  for 
24  cts.  per  pound.     Required — the  per  cent  protit.         Ans.  S3~. 

19.  A  gentleman  bought  a  house  and  lot  for  JB25U0,  and  sold 
them  for  $2000.     What  per  cent  was  his  loss?  Ans.  20. 

20.  A  merchant  bought  60  yards  of  clo'h  at  $6  per  yard,  and 
sold  the  entire  quantity  for  ^400.     What  percent  was  his  protit? 

Ans.    llj. 

21.  A  speculator  laid  out  $2000  for  land,  and  afterwards  sold 
his  land  for  !t^2800.     What  per  cent  did  he  gain?  Ans.  40. 

■    22.   Bought  cloth  for  24  cents  per  yard,  and  sold  it  for  21  cents 
per  ynrd.     What  per  cent  was  lost?  Ans.   12^. 

23.  A  merchant  bought  cloth  at  SO  .  50  per  yard,  and  sold  it  for 
^0.60  per  yard?     Required — the  per  cent  gained.  Ans.  20.- 

24.  A  man  bought  a  pair  of  tine  boots  for  85,  and  sold  them 
for  f  5.25.     What  per  cent  was  gained?  Ans.  6. 

25.  A  grocer  bought  6  sacks  of  codec,  each  containing  200  lbs., 
at  9j  cts.  per  lb.;  and  3  cwt.  of  sugar,  at  8  cts.  per  lb.:  he  sold 
the  whole  for  ^161,     What  per  cent  did  he  gain?  Ans.    I63. 

Case  3. — The  price  at  which  an  article  is  sold,  ami  the  rate  per  cent 
of  gain  or  loss  being  given,  to  find  the  original  coit. 

1.  A  man  sold  a  horse  for  890,  and  by  so  doing  i^^ained  20  per 
cen'.     How  much  did  the  horse  cost  him?  Ans.   '$75. 

Solution. — From  the  nature  of  the  question,  ll;e  selling  price  is 
120  per  cent  of  the  purchase  price.  120  p^r  cent  is  y|J.  Then 
^90  is  \~~  of  what  number? 

Proof.— 875  X  .20  =  815;  $75 +  '$15  =  $.90,  the  selling 
price. 

2.  A  man  sold  cloth  for  ^4  per  yard,  and  by  so  doin^  lost  20 
per  cent.     Required — the  original  cost.  An?.  ^5. 

Solution. — The  selling  price  in  this  example  is  evidently  equal  to 
80  per  cent  of  the  original  cost.  80  per  cent  is  jY^^.  Then  ^4  is 
yVo  o^  what  number? 

Proof.— f  5  X  •  20  =  81  ;  85  —  81  =  84,  the  selling  price. 

Hence — To  find  the  original  cost,  when  the  selling  price  and  the 
rate  per  cent  of  gaia  or  loss  are  given  : 

Obs.  6.  Add  the  gain  to,  or  suUfaci  the  loss  from  100  ;  then  mid- 
tlply  the  selling  price  by  \Q0^  and  divide  the  product  hy  the  sum  or  re- 
mainder, as  the  case  may  be. 

When  the  selling  price,  aud  thei  per  cent  of  gam  or  lots  are  given,  how  do  w» 
find  I  he  origiual  cost? 


260  COMMON    ARITHMETIC.  Scct.  XIII 

u  ■  ■  -' 

Remark. — Pupils  sometimes  think  that  if  we  find  the  percentage  on  the 
selling  price  at  the  given  rate,  and  add  it  to,  or  subtract  it  from,  the  sell- 
ing price,  the  sum  or  remainder,  as  the  case  may  be,  will  be  the  original 
cost.  This  error  may  be  avoided  by  recollecting  that  the  cost  or  purcJiase 
price  is  always  made  t/ie  basis  upan  which  the  gain  or  loss  is  calculated. 
(Obs.  1.) 

3.  By  selling  cloth  at  $5  per  yard,  25  per  cent  was  gained.  Re- 
quired— the  original  cost.  Ans.  $4. 

4.  By  selling  sugar  at  11  cents  per  pound,  10  per  cent  was 
gained.     How  much  did  it  cost?  Ans.   10  cts.  per  lb. 

5*  A  grocer,  by  selling  tea  at  81  per  lb.,  gained  llj  per  cent. 
How  much  did  it  cost  him?  Ans.  $0  ,  90  per  lb. 

6.  By  selling  cloth  at  81  .  50  per  yard,  25  per  cent  was  lost.  Re- 
quired— the  original  cost.  Ans.  82  per  yd. 

7.  A  bookseller  sold  a  lot  of  books  for  8400,  and  by  so  doing 
gained  33^  per  cent.     How  much  did  they  cost  him? 

Ans.  8300. 

8.  A  merchant  bought  a  lot  of  damaged  goods  for  81000,  which 
was  16  J  per  cent  less  than  cost.     Required — the  cost. 

Ans,  81200. 

9.  A  man  bought  two  horses,  paying  the  same  sum  for  each. 
He  afterwards  sold  one  of  them  for  8180,  which  was  10  per  cent 
loss,  and  the  other  at  15  per  cent  profit.  For  how  much  did  he  sell 
the  latter  horse?  Ans.  8230. 

10.  A  man  sold  a  piece  of  land  for  82000,  by  which  means  he 
lost  16|  per  cent.  He  tried  at  first  to  sell  it  at  12^-  per  cent  profit. 
For  how  much  did  he  try  to  sell  it?  Ans.  82700. 

Article  4.     STOCKS,  BROKAGE,    DUTIES,  and  TAXES. 

STOCKS. 

Obs.  1 .  Bt/  Stock  is  meant  the  capital  of  Trading  Companies, 
Banks,  Mail-road  and  Insurance  Companies,  Manvf actor ies,  t&c; 
also,  the  funds  of  governmnnt,  state  bonds,  d'c.  It  is  generally  di- 
vided into  a  certain  number  of  parts,  called  S  hares.  A  share 
is  generally  valued  at  8100,  and  owners  of  shares  are  called 
Stock- holders. 

Obs,  2.  Stock  has  two  values  :  a  nominal,  and  a  real  value. 
The  nominal  value  is  the  original  cost,  ov  price  paid  for  a  share.  The 
real  value  is  the  price  at  which  a  share  can  be  sold,  which  varies  at 
different  times. 

Obs.  3.     When  stock  sells  for  its  nominal  value,  it  is  said  to  be 

What  error  do  pupils  sometimes  fall  into  with  respect  to  questions  of  this 
kind?  How  miiy  this  error  be  avoided?  VV^hat  \^  meant  by  Stock?  How  is 
it  generally  divided?  What  sum  usually  constitutes  a  share?  What  are  the 
owners  of  a  share  called?  How  many  values  has  stock?  What  are  they? 
What  is  the  nominal  value?     The  real  value?  xt 


Art.  4.  '    '        STOCKS.  261 

at  par.  When  it  sells  for  more  than  its  nominal  value,  it  is  said  to 
be  above  yavy  or  at  an  advance  ;  and  when  it  sells  for  less  than  its 
nominal  value,  it  is  said  to  be  below  par ^  or  at  a  discount.  Par  is  a 
Latin  word,  signifying  equal. 

Remark. — Persons  who  deal  in  Stock  are  usually  called  Stock- brokers,  or 
Sioak-jobbers. 

Obs,  4.  The  variation  of  the  real  value  of  Stock  is  called  its  rise 
or  fall.  It  is  reckoned  at  a  certain  per  cent  on  its  nominal  or  par 
value, 

EXERCISES    FOR   THE    SLATE. 

1.  What  is  the  real  value  of  $1000  of  bank  stock  at  1 10  per  cent; 
that  is,  10  percent  above  par?  Ans.   1 100. 

Operation.— ^\QQO  X  I-IO  =  81 100.     Or, 
81000  X. 10  =  8100;  81000+8100  =  1100,  as  before. 

2^  What  is  the  value  of  81600  in  bank  stock,  at  94  per  cent;  that 
is,  ^per  cent  below  par?  Ans.  81410, 

,,     Operaiion.—^\mO  X  .94  =  81410.     Or, 
b:  81500  X. 06  =  890;  81500  —  90  =  81410,  as  before. 

Remark. — From  these  operations  we  perceive  that  questions  in  Stocks  are 
performed  in  the  same  manner  as  in  Percentage;  (Art.  1.  Obs.  4.)  or  as  ques- 
tions in  Profit  and  Loss,  Casel.  (Art.  3.  Obs.  2.) 

3.  Bought  8  shares  of  railroad  stock  at  12  per  cent  advance. 
How  much  did  it  cost?  Ans. 


Note. — When  the  value  of  ihe  share  is  not  given  in  these  examples,  $100  is 
understood  as  its  value.  ' 

4.  A  gentleman  bought  6  shares  of  stock  in  a  manufacturing  es- 
tablishment, at  4  per  cent  discount.     How  much  did  it  cost? 

Ans.  8576. 

5.  A  gentleman  bought  12  shares  of  stock  in  a  cotton  factory, 
at  4  per  cent  discount,  and  sold  it  at  6^  per  cent  advance.  How 
inuch  did  he  gain,  the  par  value  of  each  share  being  876? 

Ans.  892.25. 

6.  A  stock-broker  bought  20  shares  of  canal  stock,  at  2^  per 
cent  discount,  and  sold  it  at  2>\  per  cent  advance.  Required — his 
gain,  the  par  value  of  a  share  being  8100?  Ans,   120. 

When  is  stock  said  to  be  at  par?  When  above  par?  When  below  par? 
What  doe*  par  mean?  What  are  persons  who  deal  in  stock  usually  called? 
'What  is  the  variation  of  the  real  value  of  stock  called?  How  is  it  reckoned? 
How.^tjB  questions  in  stock  performed? 


262  COMMON    >E1THMETIC.  Scct.    XIII 

7.  Bought  G  shares  of  railroad  stock  at  94   per  cent,  and  sold  it 
^at  n2  per  cent.     Required — the  gain?  Ans.  $108. 

8.  Bought  9  shares  of  railroad  stock  at  106  per  cent,  and  sold  it 
at  96  per  cent.     How  much  was  lost  by  the  transaction? 

Ans.  ^90. 

BROKAOF. 

Obs,  5.  Brokage  ih'  the  operation  of  jindimj  the  percentage  on 
Bills  of  Exchange,  Bank  notes,  dx.  Persons  who  buy  and  sell 
bank  notes  are  called  Brokers. 

Obs.  6.  Bank  notes  have  two  values  :  a  nominal,  or  par  value, 
and  a  real  value.  The  nominal,  or  'par  value,  is  the  sum  named  on 
the  face  of  the  note.  The  real  value  is  the  price  at  which  the  note 
will  sell. 

The  remarks  of  Obs.  3  apply  to  bank  notes  as  well  as  to  Stock. 
The  rule  is  the  same  as  in  stocks. 

1.  What  is  the  value  of  $1000  in  bank  notes,  at  1^  per  cent  ad- 
vance? Ans.   i$  10 12. 50. 

2.  What  li  the  value  of  81500  in  bank  notes,  at  2  per  cent  dis- 
count? Ans.  $1470. 

3.  A  mer^dlant  negotiated  a  bill  of  exchange  of  $2500  with  a 
broker,  for  which  he  was  to  give  him  4|  per  cent.  How  much  did 
the  broker  receive,  and  what  sum  did  he  return  to  the  own«r? 

Ans.  The  broker  received  $110,  and  returned  •:  2390, 

4.  What  is  the  value  of  $5000  in  bank  notes,  at  f  per  cent  pre- 
mium? Ans.  $5018.75. 

5.  What  is  the  value  of  $10000  in  bank  notes  at  |-  per  cent  dis- 
count? Ans,   $9937  .  50. 

6.  A  gentleman  in  St.  Louis  exchanged  $1500  in  specie  for  Cin- 
cinnati bills,  at  5  per  cent  advance  ;  he  then  started  for  Cincinnati, 
an!  paid  for  his  passage  $15  ;  at  Cincinnati,  he  exchanged  his  bills 
for  notes  on  a  bank  in  Albany,  N.  Y.,  at  2 J  per  cent  discount;  he 
then  started  for  Albany,  and  paid  for  h  s  expenses  $45;  in  Albany 
he  sold  his  notes  for  Boston  bills,  at  \  per  cent  dis  ount:  his  ex- 
peases  from  Albany  to  l^os;!on  were  $10,  How  much  had  he  when 
he  arrived  at  Boston;  and  did  he  gain  or  lose  by  exchanging,  aad 
how  much,  supposing  the  specie  to  be  every  where  current?  ^ 

d  when  he  arrived  $1458  62. 
by  exchanging,  ;j  28.62. 


,  S  He  had 

Ans.      {  n    •       ^ 

(  (jamed 


DUTIES. 

Obs.  7,     Duties  are  sums  of  money  required  Jyy  government,  to 

Wh;it  is  Brokafire?  What  are  persons  who  buy  and  sell  bank  nott^s  CHlif>d? 
How  many  values  have  bank  notes?  What  is  the  nominal,  orpar  value?  The 
real  value?     What  is  the  rule  in  Brokage? 


Art.  4.  DUTIES,  263 


he  paid  on  the  imporlation  of  goods.  They  are  of  two  kinds  :  <S/;(?- 
ci/ic  rtHvl  Ad  Valoram,  The  income  arisin-^  from  duties,  &c.,  is 
called  tiEVEKUE  ;  and  a  table  of  the  duties  is  called  a  Tariff. 

Case    1. — Specific  Duties* 
Obs.  8.     A  Specific  Duly  is  a  certain  sum  imposed  upon  a  ton, 
pound,  yard,  gallon,  hogshead,  (fee,  without  regard  to  the  value  of 
the  article. 

Remark  1. —  Accordinof  to  law,  certain  deductions  a'--  maio,  c;i!ied  tare, 
draft,  leukairt,  &c.,  before  specific  d  at  res  are  rai  posed. 

2. — Tare  is  Die  allowance  made  jor  ihe  weight  of  the  box,  cask,  or  what- 
ever contains  the  article  on  which  the  duly  is  paid. 

Draft,  or  Tret,  is  the  allowance  made  on  the  weight  of  ihe  goods  for 
waste  or  refuse  matter. 

Lfak'ige  is  an  allowance  for  waste  of  liquors  in  casks,  &-c.  It  is  usnahy 
about  2  per  cent. 

3. — For  breakage,  10  per  cent  is  allowed  on  all  ale,  brer,  and  porter,  in 
bottles,  and  5  per  cent  on  all  other  liquors  in  bottles  ;  or  the  i!r«porier  may 
hnve  the  duties  compiled  on  the  actual  quantity  by  tale,  if  he  so  chooses 
at  the  time  of  entry. 

4.  —  At  the  Custom  Houses,  coininon  sized  bottles  are  estimated  to  contain 
22^'  gallons  per  riozen. 

5. —  he  er.tire  tceight  of  any  lot  of  poods,  together  with  the  box,  or  what- 
ever contains  the  jjouds,  is  called  th«  grans  weight.  The  weight  or  th"  goods 
a'one,  after  -'sll  dedoctir)n9  for  tare,  draft,  ifcc,  is  called  tiie  neat  loeiuht. 
(Sect.  IX.  Art.  2.  Obs.  21.  Rem.  2.) 

1.  What  is  the  specific  duty  on  30  hhds,  of  sugar,  at  2\-  cents 
per  pound  ;  the  gross  weight  being  9  cwt.  2  qrs.  each  ;  the  draft 
4  lbs.  per  hhd.,  and  the  tare  12  per  cent? 

Operai'on. 
9  cwt.  2  qis.  =  950  lbs.  in  each  hhd. 

30 


28500  lbs.  in  30  hhds. 
30  X  4=  120  lbs.  draft. 


28380 
28380  X  .12=       34:  6  lbs.  tare. 


2)24974  lbs.  neat  weight. 
.021  duty  per  lb. 


49948 
12487 

Ans.  $624. 3o 


264  COMMON    ARITHMETIC.  ScCt.    XIII 

In  tliis  operation  we  notice  two  considerations  : 

1st.  The  tare  is  oomputed  on  the  remainder,  after  the  draft  is 
deducted. 

2nd.  The  duty  is  computed  on  the  neat  weight,  or  the  remainder 
after  all  deductions  are  made. 

Note. — In  casting  tare,  any  remainder  which  does  not  exceed  half  a  pound, 
is  not  reckoned  ;  but  if  it  exceeds  half  a  pound,  it  is  reckoned  a  pound.  By 
remainder,  in  this  case,  is  meant  the  decimals.     Hence — 

To  lind  the  specific  duty  on  any  article  of  merchandise  : 

Obs.  9.  First  deduct  the  draft,  tare,  leakage,  (&c.,  from  the 
given  article,  and  then  multiply  the  remainder  by  the  duty  per  pound, 
gcdlon,  yard,  dc, 

2.  What  is  the  specific  duty  on  10  hogsheads  of  wine,  at  12 
cents  per  gallon,  deducting  2  per  cent  for  leakage? 

Ans.  $74,088. 

3.  What  is  the  duty  on  5  gross  of  porter,  at  1 5  cents  per  gallon, 
10  per  cent  being  allowed  for  breakage?  Ans.  ^300. 

4.  What  is  the  duty  on  30  casks  of  nails  ;  weight,  140  lbs.  per 
cask,  at  4\  cts.  per  pound,  allowing  1  lb.  per  cask  for  draft,  and  3 
per  cent  for  tare?  Ans.  $182.02. 

5.  What  is  the  specific  duty  on  6  hhds.  of  sugar,  at  3  cents  per 
pound,  gross  weight  as  follows :  the  first,  6  cwt.  3  qrs.;  2nd,  7  cwt. 
1  qr.;  3d,  4  cwt.;  4th,  5  cwt.  3  qrs.;  5th,  6  cwt.  1  qr.  20  lbs;  6th,  8 
cwt.  15  lbs.:  allowance  for  draft,  4  lbs.  per  hhd.,  and  for  tare  12 
per  cent?  Ans.  .*^  100.  02. 

Case  2. — Ad  Yaloram  Duties. 
Obs.   10.    Ad  Valoram  Duties  consist  of  a  certain  per  cent  of 
the  actual   c'ost  of  the  goods    in  the    country  from  ivhich  they  were 
brought. 

Ad  Valoram  is  a  Latin  term,  signifying  according  to  value. 

1.  What  is  the  ad  valoram  duty  on  a  quantity  of  broadcloth,  in- 
voiced at  $2000  in  London,  at  12  per  cent?  Ans.  $240. 

Suggestion. — When  the  duty  is  imposed  on  the  actual  cost  of  the 
goods,  of  course  no  deductions  are  to  be  made.     Then  12  per  cent 

What  are  Duties?  How  many  kinds  of  duty  are  there?-  What  is  Revenue? 
What  is  Tariff?  What  are  Specific  Duties?  What  deductions  are  made  be- 
fore Specific  duties  are  imposed?  What  is  Tare?  Draft?  Leakage?  What 
per  cent  is  allowed  on  liqi'ors  in  bottles,  for  breakage?  How  much  are 
common  sized  bottles  estimated  to  contain?  What  is  gross  weight?  Net 
weight?  Upon  what  is  tare  computed?  The  duty?  How  do  we  find  the 
specific  duty  on  an  article  of  merchandise?  What  are  Ad  Valoram  duties? 
What  does  the  term  ad  valoram  mean?  Are  any  deductions  to  be  made  on 
ad  valoram  duties?    Why  not? 


Art.  4.  TAXES.  265 

■  of  $2000,  is  S2000  X  •  1 2  =  $240.     It  will  be  perceived  that  the 
operation  is  the  same  as  in  percentage.     (Art.  1.  Obs.  4.) 

Remark. — An  Invoice  is  a  written  statement  of  the  goods  in  qaestion, 
with  the  prices  of  the  articles  annexed. 

2.  What  is  the  duty  on  a  quantity  of  books,  invoiced  at  $800,  at 
10  per  cent,  ad  valoram?  Ans.  $80. 

3.  What  is  the  duty  on  a  quantity  of  wine,  invoiced  at  $1000, 
at  165:  per  cent,  ad  valoram?  Ans.  $162.50. 

4.  What  is  the  duty  on  a  quantity  of  jewelry,  invoiced  at  $500, 
at  1 8  per  cent,  ad  valoram  ?  Ans.  $90, 

5.  A  merchant  bought  a  quantity  of  silks  in  Canton  for  1200 
tales  ;  upon  importing  them  the  duty  was  1 5  per  cent  ad  valoram. 
Required — the  duty,  the  tale  being  estimated  at  $1  .  48. 

Ans.  $266  .  40. 

TAXES. 

Obs.  11.  Taxes  are  sums  paid  bi/ the  people /or  the  support  or 
benefit  of  the  Government,    County,  District,  <^c. 

Obs.  12.  Taxes  are  levied  either  on  the  person  or  property  of 
the  citizens.  When  the  tax  is  levied  on  the  person,  it  is  called  a 
poll  tax,  and  is  usually  a  specified  sum.-f 

When  the  tax  is  levied  on  the  property,  it  consists  of  a  certain 
per  cent  of  the  value  of  the  entire  property  possessed  by  each  tax- 
able person. 

Remark. — Property  is  of  two  kinds  :  Real  Estate  and  Personal  Property 
Real  Estate  denotes  property  that  is  fixed  :  as,  lands,  houses,  &c.  Personal 
Property  denotes  property  that  is  moveable  :  as,  horses,  cattle, carriages,  notes, 
tools,  &c. 

Obs.  13.  The  first  thing  necessary  to  be  obtained  in  levying  a 
tax  is  an  exact  inventory,  or  written  account,  of  all  the  taxable  pro- 
perty, both  real  and  personal,  in  the  State,  County,  or  District  &c., 
where  the  tax  is  to  be  levied,  together  with  the  value  of  the  entire 
property  owned  by  each  individual  who  is  to  be  taxed,  and  the 
number  of  polls. 

An  Inventory  is  a  list  of  articles. 

1 .  A  certain  district  was  taxed  $1850.  The  district  contained  50 
polls,  which  are  assessed  at  $1  each;  and  the  whole  amount  of  tax- 


How  do  we  find  the  ad  valoram  duty  on  an  article?  What  is  an  invoice? 
What  are  taxes?  How  levied?  What  is  a  poll  tax?  When  a  tax  is  levied  on 
property,  of  what  does  it  consist?  Of  how  many  kinds  is  property.?  What 
is  Real  Estate?  Personal  Properiy?  What  is  the  first  thi  ng  necessary  to  b» 
obtained  in  levying  a  tax? 

tin  sttpe-fitaies tireffe  fs  no i»t:)il  rapc^.  ^'^    *"Wii%i  aii^iMMi. yA  tfWt  mdJt «  wttik^ 


266  OOMMOK    ARITHMETIC.  Scct.    XIII 

able  property,  both  personal  and  real,  in  the  district,  was  valued  at 
$120000.  Required — the  tax  on  a  dollar,  and  also  the  tax  of  an 
individual  whose  real  estate  is  worth  ^2500,  and  whose  personal 
property  is  worth  ^800,  and  who  pays  for  1  poll? 

Ans.  Tax  on  $1  is  1^  cts.;  on  the  individual  $50,50. 

p"!*^  Operation, 

$  1 850  1 20000)  1 800 .  00 ( .  0 1  tVoVVo  =  1  \  cts.  on 

60  poll  tax.  1200  00  [the  doll. 

$1800  tax  on  property.  600000 

$2500  real  estate.  $3300 

800  personal  property.  .01^ 


$33J0  taxable  property.  $3300 

1660 


60 +  $1  =  $60.50. 

As  the  poll  tax  is  a  specified  sum  on  each  individual,  it  is  evi- 
dent that  this  must  be  deducted  from  the  tax  to  be  raised  before  cal- 
culating the  tax  on  the  property.  By  this  means  we  find  the  tax  on 
the  property  to  be  $1800.  Then  if  $120000  pays  $1800  tax,  $1 
will  pay  $1800 —  $120000=  1^  cts.  Also,  if  a  man's  real  es- 
tate is  worth  $2500,  and  his  personal  property  worth  $800,  his  en- 
tire taxable  property  is  worth  $2500  -|-  $800  =  $3300;  and  $3300 
X  .  li  =  $49  .  60,  and  $1  for  his  poll  tax  makes  his  entire  tax  to 
be  $60 .  60.     Hence— 

To  assess  a  Statt ,  County,  or  other  tax,  we  have  this 

RULE. 

1.  Find  the  tax  on  all  the  polls,  (if  any)  at  the  given  rate,  and 
subtract  this  from  the  whole  tax  to  be  raised.  Divide  the  remainder 
hf  the  whole  amjount  of  taxable  property  in  the  State,  County,  Dis- 
trict, <&c.,  and  the  quotient  will  be  the  tax  on  one  dollar. 

II.  Multiply  the  value  of  each  man's  property  by  the  tax  on^\,  and 
the  product  will  be  the  tax  on  his  property,  to  which  add  his  poll 
tax,  and  the  sum  will  be  his  total  tax. 

Peoof. — Add  together  the  taxes  of  all  the  individuals  taxed  ;  if 
the  svm  is  equal  to  the  whole  tax  assessed,  the  work  is  correct. 

2.  A  town  was  taxed  $600.  It  contained  40  polls,  valued  at 
$0 .  76  each,  and  the  value  of  the  taxable  property  was  $23500. 

What  muit  be  done  before  the  tax  is  lerird  on  the  property?  Why  »o? 
What  ia  th«  rule  for  u«e»£Aog  taxw?    Tho  proof? 


Art.  4y< 


TAXE 


267 


What  per  cent  was  the  tax,  and  what  is  the  tax  of  an  individual  who 
pays  for  3  polls,  and  whose  property  is  worth  61575? 

.  i  1  ax  on  ^  I  is  2  cents. 

^"^-       I  Individual's  tax,  833.75. 
3.  A  company  was  taxed  $3l3!k     The  value  of  the  taxable  pro- 
perty was  ^1 '25000,  and  there  were  20  polls  valued  at  80  .  50  each. 
Required — the  tax  on  ^  1 . 


Ans.  2  J  cts. 


Remark. —  A«se8»ors  generally  make  a  table  containing  the  tax  on  '$1,  $2, 
&.C.,  up  to  $1000  ;  and  thus,  by  knowings  the  value  of  any  one's  property,  hi* 
tax  can  easily  ba  found.  Thus,  in  ihti  last  example,  the  tax  on  $1  being  2i 
cts.,  on  $2  it  will  be  twice  as  much,  or  5  cts.,  &c. 


Hence — the  follow  in  a: 


TABLE. 


$1  pays 

$0.02^.  iS20pays 

80.50. 

8200  pays 

$5.00. 

2  '* 

.05.  1  30  '' 

.75. 

300  '* 

7.50. 

3  " 

.071. 

40  " 

1.00. 

400  " 

10.00. 

4  " 

.10. 

50  ** 

1.25. 

500  " 

12.50. 

5  " 

.121.   60  " 

1.50. 

600  '' 

15.00. 

6  " 

.15.  i  70  " 

1.75. 

700  '' 

17.50. 

7  " 

.171. 

80  " 

2.00. 

800  *' 

20.00. 

8  " 

.20. 

90  " 

2.25. 

900  " 

22.50. 

9  « 

.221. 

100  " 

2.50. 

1000  *' 

25.00. 

10  " 

.25.  1 

4.  Find,  by  the  above  table,  the  tax  of  an  individual  whose  pro- 
perty is  valued  at  ^1426,  and  who  pays  for  2  polls,  at  $0.50  each. 

Ans.  836.65. 


Operation. 
The  tax  on  81000  is  825.00 
The       **  400  is     10.00 

The       "  20  is         .50 

The       "  6  is         .15 


And  the  tax  on  81426  is  835.65 
Two  polls  at  80.50  each  =       1 .00 


1 426= 1000-f  400+20+6. 
Therefore,  by  finding  the  tax 
on  each  of  these  in  the  table; 
and  adding  them  together 
we  find  the  tax  on  the  whole. 
(Sect.  *  4.  Art.  4.  Obs.  4. 
Rem.  2.) 


Total  tax,  836.65 
5.  What  would  be  A's  tax  in  the  above  company 
1  poll,  and  whose  property  is  worth  81594? 


who  pays  for 


Ans.  840.35. 


How  do  assessors  proceed  in  levying  a  tax?    How  do  we  find  a  '.ax  by  the 


Table? 


268  COMMON    ARITHMETIC.  ScCt.    XIII 

6.  C  pays  for  3  polls,  and  his  property  is  worth  $2875.  Required 
his  tax?  Ans.  $73.37^. 

7.  D  pays  for  1  poll,  and  his  property  is  worth  .$882.  Required 
his  tax?  Ans.  $22.55. 

8.  E  pays  for  2  polls,  and  for  property  worth  ^27740.  Re- 
quired his  tax?  Ans.  '8694.50. 

9.  F  pays  for  4  polls,  and  for  property  worth  !$48450.  Required 
his  tax?  "  Ans,  81213.25. 

10.  G  pays  for  3  polls,  and  for  property  worth  ^1^3001 2.  Required 
his  tax?  Ans,  $751,80, 

1 1.  H.  pays  for  4  polls,  and  for  property  valued  at  $12021.  Re- 
quired his  tax?  Ans.    $302,523. 

12.  Prove  by  the  8  preceding  examples,  that  the  result  in  exam- 
pi  j  o  i:jCtiVvCC. 

Article  5.     Interest. 

Obs.   1.     Interest  is  the  sum  paid  for  the  use  of  money  ^ 

It  is  computed  by  ^?er  ce??/a_j7e.'  that  is,  so  many  dollars  are  paid 
for  the  use  of  $100  for  1  year;  and  in  the  same  proportion,  for  a 
sum  greater  or  less  than  $100,  or  for  a  longer  or  shor  er  time  than 
a  year. 

Obs,  2.  The  sum  on  which  the  interest  is  paid  is  called  the 
Principal, 

The  interest  is  generally  a  certain  per  cent,  of  the  priiuipal.  The 
per  cent;  on  $100  for  1  year  is  called  the  Rate. 

The  sura  of  the  principal  and  interest  is  called  the  Amount. 

Illustration. — If  I  borrow  $100  for  1  year,  and  agree  to  pay  6 
per  cent,  for  the  use  of  it,  at  the  end  of  the  year,  I  must  pay  back, 
not  only  the  $100  which  I  borrowed,  but  6  per  cent,  of  it, 
or  $6  also,  making  in  all  $106.  In  this  case  $100  is  the  principal; 
$6  the  interest,  6  per  cent,  the  rate;  and  $106  the  amount. 

Obs.  3.  Per  annum  signifies  by  the  year.  Thus,  6  per  cent. 
per  annum,  signifies  that  $6  is  paid  for  the  use  of  $100  for  1  j-ear. 

Remark. — When  the  time  is  not  mentioned  with  the  rate  per  cent.,  a  year 
is  always  understood. 

Obs.  4.     Legal  interest  is  the  rate  per  cent,  established  by  law. 

What  is  interest?  How  is  it  computed?  What  iis  the  principal?  Tlip  in- 
terest? The  rate'?  The  amount?  In  the  illustration  given,  which  is  the 
principal?  The  interest?  The  rate?  Thef<monnt?  What  doe«  per  annum 
signify?  Give  an  example.  When  the  time  is  not  mentioned,  with  the  rate 
pertent^j  what  is  understtfotl?    What  is  legal  interest?  _T 


Art.  5.        ^-  INTEREST.  269 

It  varies  in  different  countries,  and  also  in  the  different  States.    Thus, 
it  is 

6  per  cent,  in  all  the  New  Enp^land  States,  in  New  Jersey,  Penn- 
sylvania, Delaware,  Maryland,  Virginia,  North  Carolina,  Tennessee, 
Kentucky, Ohio,  Indiana,  Illinois,  Missouri,  Arkansas,  District  of  Co- 
lumbia, and  on  debts  or  judgments  in  favor  of  the  United  States. 

7  jjer  cent,  in  New  York,  South  Carolina,  Michigan,  Wisconsin, 
and  Iowa, 

8  per  cent,  in  Georgia,  Alabama,  Mississippi,  Te.xas,  an  d 
Florida. 

5  per  cent,  in  Louisiana, 

Remark.  1.  When  the  rate  per  cent,  is  not  mentioned  in  transacting  busi- 
ness, the  legal  rate  is  always  uu'ierstood. 

2.  Any  rate  per  cent,  higher  than  the  legal  rate  is  called  usury,  and  the 
person  who  exacts  it  is  liable  to  punisimieat.  If  the  parties  agree,  however, 
any  rale  less  than  the  legal  rate  nmy  be  taken. 

Obs.  5.  The  learner  will  observe  tliis  difference  between  per 
centage  and  interest:  in  interest,  time  is  considered;  in  per  centa<je  it 
is  not. 

Interest  is  of  two  kinds  :     Simple  and  Compound. 
SIMPLE  INTEREST. 

Obs.  6.     Simple  Interest  is  the  interest  on  the  principal  only , 

Case  1.  To  calculate  interest  for  any  numher  of  years,  at  any 
rate  per  cent, 

Ex.  1.  What  is  the  interest  of  $450  for  3  yrs.  at  5  per  cent?  — 
The  amount? 

The  interest  is  5  cents  for  100  cents,  or  Operation. 

$1   for  1  year.  Therefore,  the  interest  of  8450 

$450  for  I  year  is  $450 X. 05  =  $22.50;  ,05 

and  for  3  years  it  is  3  times  as   much  or  

$67,50.  The  sum  of  the  interest  (-^^67.50)  $22.50  int.  for  1  yr. 

and  the  principal  ($450)  is  the  amount.  3 

(Obs.    2.)  

Ans.  $67.50  int.  for  3  yrs. 
450.00  principal. 


;  "^  V  Ans.  $517.50  amount. 

Hence — to  find  the  interest  of  any  sum  of  money,  for  any  num- 
ber of  years,  at  any  rate  per  cent. 

What  is  the  iegsl  rate  i  n  most  of  the  United  States?  When  the  rate  per 
cent,  is  not  mentioned  in  transacting  business,  what  is  understood?  What  ii 
usury?  What  is tlie  difference  between  per  centago  and  interest?  How  i» 
interest  divided?  What  issimple  interest?  How  do  we  find  the  interest  of 
any  sunVj'Cor  any  number  of  years,  at  any  rate  per  cent? 


270  COMMON    ARITHMETIC.  '  Scct.    XlU 

Obs.  7.  MaWply  the  principal  by  the  rate  per  cent,  expressed  deci- 
mally, this  luill  he  the  interest  for  1  gear. 

Multiply  the  interest  for  1  year  by  the  number  of  years  ^  if  the  time  is 
longer  than  1  year. 

Remark.     To  finJ  the  amount ;    Aid  the  principal  to  the  interest. 
EXERCISES    FOR    THE    SL4TJC 

1.  What  is  the  interest  of  Si 20.50  for  two  years  at  6  per  cent? 
At  8  per  cent.?     At  7  per  cent.? 

Ans.  in  order.     $14.46,  $19.28,  $16.87. 

2.  What  is  the  interest  of  $640.20  for  5  years  at  4  per  cent.? 
At  5  per  cent.?     At  7j  per  cent.? 

Ans.  in  order.     $128.04;  $160.05;  $240,075, 

3.  What  is  the  interest  of  $300  for  7  years,  at  12^  per  cent.? 

Ans.  $262.50. 

4.  What  is  the  interest  and  amount  of  $800  lor  12  years,  at  8  J 
per  cent.? 

Ans.  Int.  $840;  Amt.  $1640. 

5.  What  is  the  interest  and  amount  of  $1000  for  5  years,  at  4 J 
per  cent.? 

Ans.  Int.  $243.75;  Amt.   $1243.75. 

6.  What  is  the  interest  and  amount  of  $575  for  6  years,  at  4  per 
cent.? 

Ans.  Int.  $138;  Amt.  $713. 

7.  What  is  the  interest  and  amount  of  $860  for  10  years,  at  7-^ 
per  cent.? 

Ans.  Int.  $645;  Amt.  $1505. 

Case  2.     To  calculate  interest  for  years  anfJ  months,  at  6  per  cent. 

1.  What  is  the  interest  of  $400  for  1  year  and  6  months,  at  6  per 
cent  1 

At  6  per  cent.,  the  interest  of  $1  for  1  year  is  Operation. 

6  cents.     Then  as  there  are  12  months  in  a  year  $400 

the    interest  lor   1   month  is  6  -^  1 2  =  y ^  =  ^  u  .09 

cent  or  1    cent  for  every  2  months.     Then  if  it  

is  1  cent  for  (  very  2  months,  for  8  months  it  will       Ans.  $36.00 
be  4  cents;  for  10  months,  5  cents  ;  for  6  months,  3   cents,    (fee; 
that  is,  the  interest  of  1  for  any  numher  of  months,  is  half  as   many  ' 
cents  as  there  are  months  in  the  given  time. 

Therefore,  in  the  above  example,  1  yr.  6  mo.  =  18  mo  ,  anl  the 
interest  (f  -^l  for  18  mo.  is  9  cts.;  and  the  interest  of  $400,  is  .09 
X400==$36.     Hence— 


How  do  we  find  the  amount?     To  what  i«  th*  iRti^re-si  of  $1  for  any  number 
of  months  equal?     Show  why  thia  is  correct. 


Art.  5.  iiCTEREsr,  271 

To  find  the  interest  of  any  sum  for  years  and  months,  at  6  per 
cent. 

Obs.  8.     Call  kcdf  the  number  of  moTiths  cents;  and  multiply  the 
prinzipal  by  this,  expressed  decimally, 

2.  What  is  the  interest  of  $350  for  3  yrs.  8  mo.,  at  6  per  cent.? 

Ans.  $77. 
3  What  is  the  interest  and  amount  of  $1000  for  5  yrs.? 

Ans.  Int.  $300  ;  Amt.  $1300. 
When  the  rate  per  cent,  is  not  mentioned   in   these   examples,  6 
per  cent  is  understood. 

4.  What  is  the  interest  of  $1200  for  4  yrs.  7  mo? 

Ans.  $330. 

5.  What  is  the  interest  of  $750  for  15  yrs.  Ans.  $675. 

6.  What  is  the  interest  of  $225.50  for  2  yrs.  4  mo.? 

Ans.  $31.57. 

7.  Required  the  interest  of  $120  for  7  mo.?  Ans.  $4.20. 

8.  Required  the  interest  of  $284.66  for  3  yrs.  9  mo.? 

Ans.  $64,048 

9.  What  is  the  amount  of  $427.86  for  7  yrs.  5  mo.? 

Ans.  $618,257  _| 

10.  What  is  the  amount  of  $368.27  for  3  yrs.  3  mo. 

Ans.  $440,082.+ 

11.  What  is  the  amount  of  $218.75  for  10  yrs.  2  mo? 

Ans.  $352,187.+ 

12.  What  is  the  amount  of  $768»25  for  4  yrs.  10  mo.? 

Ans.  $991,042.+ 

Case.  f.     To  calciUate  the  interest  on  any  sum,  for  any  number  of 
days  at  6  per  cent. 

1.  What  is  the  interest  of  $80  for  24  days? 

Operation.  At   6   per  cent  the  interest    on   $1    for    1 

$80  month   is  ^  a  cent;   (solution  of  Ex.  1.  Case  2.) 

.004  In   interest  30  days  are  reckoned  as  a  month  ; 

therefore  the  in  erest  of  $1  for  1  day  is  $0,005 

Ans.  $0,320  -^  30  =  g-^  —\  of  a  mill;  or  it  is  1  mill  for  every 

6  days.  Then  as  it  is  1  mill  for  every  6  days,  for  21  days  it  would 
be  3^  mills;  for  12  days,  2  mills  ;  for  15  days,  ^\  mills,  <fec. ;  that 
is,  the  interest  of  %\  for  any  number  of  days,  is  one  sixth  as  many 
mills  as  there  are  days  given. 


How  do  wo  find  the  intereot  of  any  sum  for  y©«rs  and  months,  at  6  per 
cent.?  To  what  i«  tho  interest  of  $1  for  any  number  of  days  equal!  Show- 
why  tbii  ia  corr«ct. 


273  COMMON    ARITHMETIC.  ScCt.   XlII 

Therefore,  in  the  above  example  the  interest  of  81  for  24  days  is 
4  mills,  and  the  interest  of  ^^80  is  .004X80  =  80.32.     Hence— 

To  find  the  interest  on  any  sum  for  any  number  of  days: 

Obs.  9.      Call  one  sixth  of  the  number  of  days  mills;  and  multiply 
the  principal  by  this,  expressed  decimally. 

2.  What  is  the  interest  of  850  for  28  days?  Ans.  80.233.+ 

3.  What  is  the  interest  of  8120  for  20  days?         Ans.  80.40. 

4.  What  is  the  interest  of  8145  for  18  days?         Ans.  80,435. 
6.  What  is  the  interes  of  81 80  for  17  days?  Ans,  80.61. 

6.  What  is  the  interest  of  8200  for  15  days?  Ans.  80.50. 

7.  Required  the  interest  of  8260  for  12  days?  Ans-  80.60, 

8.  Required  the  interest  of  8300  for  7  days?  /ms.  80.35. 

9.  Required  the  interest  of  8900  for  5  days?  Ans.  80.75. 

10.  Required  the  interest  of  81000  for  1  day?      Ans.  80.16|, 

Case  4.     To  find  the  interest  of  any  number,  for  any  time,  at  6  per 
cent. 

Note. — This  case  is  simply  the  three  preceding  cases  united  into  one. 

1,  What  is  the  interest  of  81200  for  2  yrs.  5  mo.  18  da.  at  6  per 
cent. 

Operation.         2  yrs.  5  mo.  =  29  mo. ;  29  -r-  2  =  80.1 4^  = 

81200  80.145;  1 8  H- 6  =  3  mills.     80.146-4-80,003 

.148  =  80. 1 48,  the  interest  of  8 1  for  the  given  time, 

and  the  interest  of  81200   is    .148  X  1200== 

9600  8177.60. 
4800 
1200 


Ans.    8177.600 

Note. — We  generally  use  the  interest  of  $1  as  the  multiplier,  as  it  is  usually 
the  smallest  number  and  it  makes  no  difference  which  factor  we  use  as  the 
multiplier.     (Sect.  IV.  Art.  2,  Obs.  5:  Rem.) 

Hence — To  calculate  interest  at  6  per  cent,  we  have  the  fol- 
lowing 

GENERAL  RULE. 

I.  Reduce  the  given  numher  to  moiUhs  and  days:  then  take  half  the 
number  of  months,  and  call  it  cents;  (Obs.  8.)  end  if  there  is  an  odd 
month  call  it  6  mills,  and  count  another  mill  for  every  Q  days;  (Obs. 
9.)  this  will  give  the  interest  o/"  8 1  for  the  given  time. 

How  do  we  find  the  interestof  any  gum  for  any  number  of  days?  Which 
number  do  we  generally  use  a»  the  multiplier?  WSy?  What  i«  the  General 
Rule  for  calculating  interest  at  6  per  cent.  ?  «<  •,*•-  •  ^ . 


Art.  5. 


INTEREST. 


273 


II.  Midtiphj  the  given  nrincipnl  by  the  interest  of  $\,  ahead]/  found, 
tind  ■poird  off  as  in  multiplicaiion  of  decimals.  (Sect.  Vlll,  Art.  11. 
Rule, 

Note. — If  there  is  no  old  mouth,  and  th3  luimbjr  of  days  bo  less  than  6,  a 
cipht^r  must  be  put  in  the  place  of  mills. 

For  the  coiivenieace  of  the  learner  we  subjoin  the  following 

TABLE 

By  which  to  find  the  number  of  days  from  any  given  date  to  any 
required  date;  or  to  find  any  required  date  from  a  given   date. 


l.id'x- 

Jan. 

Feb. 

Mar. 

■ 
Apr-. 

May. 

June. 

July 

Aug. 

Sept. 

Oct. 

Nov. 
305 

Dec 

1  i 

1 

32 

60 

91 

121 

152 

182 

213 

244 

274 

335 

2  i 

2 

33 

61 

92 

122 

153 

183 

214 

245 

^275 

306 

336 

3  i 

3 

34 

62 

93 

123 

154 

184 

215  1  246 

276 

307 

337 

4 

4  35 

63 

94 

124 

155 

185 

216  1  247 

277 

308 

338 

5 

5 

36 

64 

95 

125 

156 

186 

217  :  248 

278 

309 

339 

6 

6 

37 

65 

96 

126 

157 

187 

218  i  249 

279 

310 

340 

7 

7 

38 

66 

97 

127 

158 

188 

219  1  250 

280 

311 

341 

8 

8 

39  ' 

67 

98 

128 

159 

189 

220  j  251 

281 

312 

342i 

9. 

9 

40 

68 

99 

129 

160 

190 

221  :  252 

282 

313 

3431 

10 

10 

41 

69 

lOO 

130 

161 

191 

222  253 

283 

3l4 

344! 

11 

11 

42 

70 

lOl 

131 

162 

192 

223  j  354 

284 

315 

345| 

12 

12 

43 

71 

102 

132 

163 

193 

224  i  255 

285 

316 

346 

13 

13 

44 

72 

103 

133 

164 

194 

225  \   256 

286 

317 

347 

14 

14 

45 

73 

104 

134 

165 

195 

226  257 

287 

318 

348 

15 

15 

46 

74 

105 

135 

166 

190 

227  258 

288 

319 

349 

16 

16 

47 

75 

106 

136 

167 

197 

2-28  j  259 

289 

320 

350 

17 

17 

48 

76 

107 

137 

168 

198 

229  260 

290 

321 

351 

18 

18 

49 

77 

loa 

138 

169 

199 

230  261 

291 

322 

3521 

19 

19 

5) 

78 

109 

139 

170 

200 

231  1  262 

292 

323 

3531 

20 

20 

51 

79 

no 

140 

171 

201 

232  1  263 

233 

324 

354, 

21 

21 

52 

80 

HI 

141 

172 

2i)2 

233  '   264 

294 

325 

355! 

22 

22 

53 

81 

112 

142 

173 

203 

234  i  265 

295 

326 

356 

23 

23 

54 

82 

113 

143 

174 

204 

235  {  266 

296 

327 

357 

24 

24 

55 

83 

114 

144 

175 

205 

236  !  267 

297 

828 

358 

25 

25 

56 

84 

115 

145 

176 

206 

237  I  268 

298 

329 

359 

26 

26 

57 

85 

116 

146 

177 

207 

238  269 

299 

33  • 

360 

27 

27 

58 

86 

117 

147 

178 

203 

239  1  270 

300 

331 

361 

28 

23 

59 

87 

1I8 

148 

179 

209 

240  '  271 

301 

332 

362 

29 

29 

60 

88 

119 

149 

180 

210 

241  ,  272 

302 

333 

363 

30 

30 

«* 

89 

120 

150 

181 

2U 

242  i  273 

303 

334 

364 

31 

31 

♦* 

90 

»*« 

151 

#*» 

212 

243  j  *** 

304 

»** 

3!55 

This  table  may  be  applied  to  the  solution  of  t^/o  problpms. 
Problem  I,     To  iitid  the  time  between  anv  two  dates: 


How  do  we  find  the  time  between  two  dates  by  the  tablo? 


13a 


274  COMiMON    ARITHMETIC.  Scct.    Xlil 

Obs,  10.  Find  the  day  of  the  month  in  the  index,  and  opposite  this 
in  the  column  of  the  given  months  ivill  be  found  the  day  of  the  year 
upon  which  the  different  dates  fall.  The  difference  of  these  days,  less  1 , 
will  express  the  time  between  the  two  dates. 

Required — the  time  between  March  12th  and  June  26th.  '- 

Solution. — Opposite  12,  in  the  column  for  March  is  found  71;  op- 
posite 26,  andin  the  column  for  June  is  found  177.  177 — 71  =  106; 
106  less  1  =  105.  Ans,   105  days. 

Remark  1.  In  leap  years  when  the  29th  of  Feb.  comes  between  the  dates 
the  diffrence  between  the  days  is  the  time. 

2.  When  the  time  overruns  the  end  of  th*?  year,  take  the  earlier  dute  from 
365,  and  to  the  ditrerence  add  tlie  number  found  against  the  latter  date,  less  1. 

Required — the  time  between  Jan  17th,  and  March  oth,  1848, 

Ans.  50  days. 

Required — the  time  between  Dec.  14th,  and  Feb.  9th. 

Ans.  56  days. 

Required — the  time  between  Aug.  23d,  and  Oct.  17th. 

Ans.  54  days. 

Problem  2.     To  find  any  required  date: 

Obs.  1 1 .  Find  the  day  of  the  year  as  before;  to  this  add  the  time 
between  the  given  and  required  dotes,  and  find  the  sum  in  the  table.  The 
column  in  which  the  sum  is  fov.nd  will  express  the  month,  and  opposite 
the  sum  in  the  index  uill  be  found  the  day  of  the  month,  less  1. 

A  note  was  given  May  7th,  to  run  90  days;  wiien  does  it  become 
due?  Ans.  Aug.  6Lh, 

Solution. — Opposite  7,  in  the  column  for  May  is  found  127;  127 
-f-  90  =  217  ;  217  is  lound  in  the  column  for  Aug.  and  opposite  5. 
5+1=6. 

Remark  1.  In  leap  years,  when  th'.^  29t!i  of  P'eb.  comes  between  the  dates, 
the  result  foond  in  the  trtbie  is  the  correct  date. 

2.  When  the  linje  overruns  the  end  of  the  year,  take  365  from  tlie  si>m. 
and  the  difference  will  be  the  number  to  find  in  the  table,  for  the  required 
date. 

A  note  given  Jan.  16th,  1348,  fell  due  in  2  months,  at  what  time 
did  it  fall  due?  Ans.  March  17th. 

Solution. — 2  mo.  =  60  d,  16  -|- 60=  76;  76  is  in  the  column 
for  March,  and  opposite  17. 

What  ri)««t  be  obsprved  in  Leap  Years  when  the  29lh  of  Febrimry  comes 
between  the  dates?  When  the  tirut*  overruuH  tiu-  end  of  th"  ye^ir  how  do  we 
proce»'fi'?  How  do  \vr  fiiui  a  r«vjuirfd  diite  by  tli--  t'hl'-?  VYh>-n  t!:e  29'l>  of 
Feb.  conies  between  ihe  dati'S  ho'.v  do  we  proceed?  When  the  time  overruns 
the  end  of  the  year^  how  do  we  proceed?       ,__  _     .  _  .  ^ 


Art.  5.  INTEREST.  275 

What  date  is  54  days  later  than  the  22nd  of  Dec? 

Ans.  Feb.  15th. 

A  note  was  given  Au^.  29th,  payable  in  6  months  ;  when  did  it 
fall  due?  Ans.  Feb,  26th. 

A  note  was  given  Oct.  11th,  1847,  payable  in  5  months;  when 
did  it  fall  due?  Ans.  March  10th,  1848. 

A  note  was  given  Jan.  12th,  1848,  payable  in  9  mo.;  when  did 
it  fall  due?  Ans.  Oct.  9th,  1848. 

NoTK. — In  these  examples ^e  have  calculated  30  days  to  the  month.  When 
the  exact  time  is  required,  the  surplus  time  must  be  added  or  the  result  be  di- 
minished for  Feb.  if  required.  The  exact  answer  for  the  last  example  is  Oct. 
13th  184«. 

Find  the  interest  and  amount  of  the  following  sums  at  6  per 
cent. 

2.  $140  for  1  yr.  3  mo.   12  da.  Ans.  Amt.  8150.78. 

In  the  answers  to  the  following  questions  the  amount  only  is 
given. 

3.  $84.23  for  2  yrs.  4  mo.  18  da.  Ans.  $96,274.-}- 

4.  $300  for  1  yr,  8  mo.  Ans.  $330. 
6.  $24  for  from  June  12th,  1847  to  April  26  1848, 

Ans.  $25,272. 

6.  $650  for  3  yrs.  1  mo.  6  da.  Ans.  $770.90. 

7.  $0.96  for  2  J  days.  Ans.  $0.9632. 

8.  $175  for  1  yr.  1  mo.  1  da.  Ans.  $186,404. 

9.  $280  for  2  yrs.  4  mo.   15  da.  Ans.  $319.90. 

10.  $315.25  for  3  yrs,  24  da,  Ans.  $373,256. 

11.  $420.50  for  4  yrs.  8  mo.  10  da.  Ans.  $538.94. 

12.  $742.80  for  5  yrs.  2  mo.  20  da.  Ans.  $975,544. 

13.  $896.96  for  6  yrs.  8  mo.  Ans.  $1255.744. 

14.  $1457.12  for  6  yrs.  6  mo.  6  da.  Ans.   $2026.853.-f. 

15.  $212.25  for  4  yrs.  1  mo.  2  da.  Ans.  $264,322. 

16.  $320.75  for  6  yrs.  3  mo.  9  da.  Ans.  $441,512.+ 

17.  $478.60  for  1  yr.  1 1  mo.  Ans.  $533,639. 

18.  $317  for  7  yrs.  9  mo.  16  da.  Ans.  $465,249. 

19.  $400  for  from  Jan  6th  to  May  17th.  Ans.  $408.6o6. 

20.  $700  for  7  yrs.  7  mo.  7  da.  Ans.  $1019.316, 

Required  the  interest  and  amount  of  the  following  sums  at  6  per 
cent: 

21.  $1250  for  9  yrs.  2  mo.  1  day.  Ans.  $1937.708. 

a.  When  the  interest  is  required  for  a  large  i  umber  of  years,  it 
is  generally  more  canvenien-,  to  find  the  interest  tor  the  years  sepa- 
rate (som  that fcvtbemoutlis anU days,  and  thea  add  the  two  re* 


276  COMMON    ARITHMETIC.  ScCt.    XlH 

suits  together.  Thus,  the  interest  of  ^1260  for  9  yrs.  is  Si 250  X 
,06  X  9  =  ^675;  (Obs.  7.)  and  for  2  mo.  1  da.  the  interest  is 
$1260  X  .010i-=  ^12:708;  and  i6/5  + $12,708  =^  .^687.708  as 
the  intei^est  of  $1250  for  9  yrs.  2  mo.  1  da.;  and  the  amount  is 
$687,708 +  $1250  =  $1937.708. 

22.  $590  for  9  yrs.  6  mo;  28  da.  Ans.  $92B.C02. 

23.  $440  for  10  yrs.  7  mo.  29  do.  Ans,  $721,526. 

24.  $111.11  for  11  yrs.  11  mo.   11  da  Ans.  $189,647, 

25.  $1200  for  11  yrs.   10  mo.  23  da.         •        Ans.  $2066.60. 

26.  $1600  for  14  yrs.  7  mo.  17  da.  Ans.  $2816.75. 

27.  $319.90  for  15  yrs.  8  mo.  26  da,  Ans.  $621,992, 

28.  $975.50  for  20  yrs.  10  mo.  14da.  Ans.   $2197.151.— 

29.  $600  for  40  yrs.  6  mo.  24  da.  Ans.  $1717. 

30.  ^1000  for  60  yrs.  12da.  Ans.  $4602. 

Case  5.  To  jind  the  interest  of  any  sum,  for  any  time,  at  any  rate 
jper  cent. 

FIRST    METHOD. 

Remark.  We  have  now  considered  how  Interest  is  computed  at  6  per  cent., 
but  it  often  happens  that  we  wish  to  find  the  interest  on  sums  of  money,  at  oth- 
er rates  per  cent,  than  6. 

The  interest  of  $1  for  1  year  at  6  per  cent,  is  6  cents; 

At  1  per  cent,  it  is  1  cent,  or  i-  of  6  per  cent, ; 

At  7  percent,,  it  is  7  cents,  or  7  times  1  per  cent.; 

At  3  per  cent,  it  is  3  cents,  or  3  times  1  per  cent;  &c.     Hence-*- 

To  find  the  interest  of  any  sum,  for  any  time,  at  any  rate  per 
cent, 

Obs.  10,  Find  the  interest  at  6  per  cent,  as  usual;  divide  this  by  6 
and  the  residt  will  be  the  interest  at  1  per  cent,  then  midtiply  the  inter- 
est at  1  per  cent,  by  the  given  rate.'*' 

Rkmark.  —  If  the  learner  chooses,  he  can  multiply  the  interest  at  6  per  cent, 
by  the  given  rate,  and  divide  the  product  by  6,   as  it   will   always   produce  the 


When  the  interest  is  required  for  a  large  number  of  years,  how  do  we  pro- 
ceed? How  do  we  compute  interest  at  other  rates  per  cent.  ti)an  6?  Show 
why  this  rule  is  correct.  By  what  other  ri.etho'd  can  the  interest  be  found? 
Demonstrate  this  rule.  :^- 

*Wlien  the  rate  (ler  cent,  is  3,  we  may  multiplv  the  principal  by  \  the  nnniVer  of  months  ; 
when  the  rate  ))er  cent,  is  4,  we  may  multiply  the  principal  ly  s^  the  number  of  months  ; 
w)ien  the  rate  percent,  is  8  we  may  multipl.if  byf  thenuiriber of  n  onthp*  and  when  tlie  rate 
per  rent,  is  9  vve  may  multiply  Ify*^  tie  number  of  month?;  and  in  ea^-h  cape  ?»oint  o^F  two 
decimals.     Those  rules  Tvilj  only  apuly  however,  when  the  timfl  caii  be  iTdtu'e;!  to  niontf!^ 

-DEMO-sTiiATipN  — 3  •  0'- cciu:,  i-3   i.    rs..for  ICO  rents  for  l^!n^^-!.^;  »:;;<•;  is  3  divided   by 
r2  =  iof  a  cent  per  montTi;  4'i)er  c>.ni  is"!  of  a  cent  noV  fTiufitii;  8  (icr  i*enT.'is  f  "bf'k'  cent" 
per  month ;  and  9  per  cent .  ig  f  of  a  cent  per  month ,    Heticc,  the  rules  are  evident' 


Art.  5.  INTEREST.  277 

same  result  and  by  this  means  fractienal  numbers  may  often  be  avoided  in  the 
opera  ion. 

1.  What  is  the  interest  of  $400  for  2  yrs.  6  mo.  18  da.,  at  4  per 
C(mt?  Ans.  ^^40.80. 

Solution. — 8400  X  •  153  =  861 .20.  the  interest  at  6  per  cent: 
$61 .20 -r- 6  =  $10.20,  the  interest  at^l  per  GCiTi.;  and  $10.20  X 
4  =::!  $40. 80  the  interest  at  4  per  cent. 

Find  the  interest  and  amount  of  the  following  sums  : 

2.  $15.30  for  9  mo.,  at  7  per  cent.  Ans.  Amt.  $16. 103. 

3.  $40  for  1  yr.  6  mo.,  at  8  per  cent.  Ans.  $44.80. 

4.  $120.60  for  18  da.,  at  9  per  cent.  Ans.  $121,142. 
5  $2f00  for  2  yrs.  4  mo.  18  da.  at  3  per  cent. 

Ans.  $214.30. 

6.  $250.76  for  1  yr.  8  mo.  24da„  at  4^  per  ceat. 

Ans.  $270,308. 

7.  $300  for  3  yrs.  2  mo,  12  da.,  at  2 J  per  cent. 

Ans.  $326.40. 

8.  $325  for  2  yrs.  6  mo.  16  da.,  at  5  per  cent. 

Ans.  $366,347. 

9.  $400.50  for  4 yrs.  11  mo.  29  da.,  at  1\  per  cent. 

Ans.  $550,604. 

10.  $525.75  for  5  yrs.  9  mo.  21  da.,  at  9  per  cent. 

Ans.  $800,585. 

1 1 .  $600  for  6  yrs.  7  mo.  1 1  da.,  at  1 1  per  cent . 

Ans.  $1036.516. 

12.  $1000  for  8  yrs.  5  mo.  15  da.,  at  12  per  cent.  5 

Ans.  $2015.  ^ 

SECOND    METHOD, 

Since  1  month  is  -^^  of  a  year;  2  months,  -^i  or  i-  of  a  year,  (fee, 
it  follows  that  the  interest  of  any  sum  for  i  month,  is  }-^  of  the  in- 
terest of  the  same  sum  for  1  year;  tor  2  months.  ^^  or  \  of  the  in- 
terest for  1  year;  for  3  months,  y^  or  \  of  the  interest  for  1  year, 
(fee.  Also,  since  1  day  is  —  of  a  month;  2  days,  /^  or  y-  of  a 
month;  3  days,  —-  or  yV  o^  ^  month,  (fee,  it  is  evident  that  the  in- 
terest of  any  sum  for  1  day  is  3-V  of  the  interest  of  the  same  sum 
for  1  month;  for  2  days,  /g-  or  yj  of  the  interest  for  i  month;  for  3  , 
days,  3%-  or  jV  o^  ^^^  interest  for  1  month;  (fee.     Hence — 

To  find  the  interest  of  any  sum,  for  any  time,   at   any   rate   per  » 
cent: 

a.'  Find  the  iriterest  for  the  years  according  to  Case  \,   Ohs.l;  then 
JiTvi  the  interest  for  the  mordhs  and  days  by  taking  aliquot  parts.  . 


273 


COMMON    ARITHMETIC. 


Sect.  XIII 


13»  Wheat  is  the  interest  of  $360  for  3  yrs.  1 1  mo.  18  da.,  at  5  per 
cent?  Ans.  '^71.40. 

Operation. 
^360 
.05 


6  mo.  =~  a  year.  2 
4  mo.  =1  of  a  year.  3 
1  mo.  =  f  of  4  mo.  4 
15  da,  =  '2  of  a  mo.  2 
3  da. =i  of  15  da.      5 


818.00  =  Int.  fori  yr. 
3 


^54.00  =  Int,  for3yr.    ] 

q  on  —  Tnf     fnv  Q  qIO*    [ 

4  mo.  ( 
1  mo.  J 


9 . 00  :=:lnt.  for  6  mo*  !  pj^  41.1  11 
6.00  =Int.  for  4  mo.  ^^-t-^-^l  -•  ^i 
1.50=  Int.  for  1  mo.  J 

75  =  Int.  for  15da  ' 

1 5  =  Int.  for 


Ans.  ^7]. 40 
Find  the  interest  and  amount  of  the  following  sums: 

14.  ^360.80  for  2  yrs.  6  mo.  15  da.,  at  6  per  cent. 

Ans.  $415,822. 

15.  .f  212.  50  for  3  yrs.  8  mo.  10  da.,  at  7  per  cent. 

Ans.  $267,454. 

16.  8400  for  5  yrs.  10  mo.  12  da.,  at  4  per  cent. 

Ans.  $493,865, 

17.  $525  for  4  yrs.  29  da.,  at  8  per  cent 

Ans.  $696,383. 

18.  $612.75  for  7  yrs.  1  mo.  20  da.  at  9  per  cent. 

Ans.   $1006.441. 

19.  $700  for  5  yrs.  10  mo.,  at  5  percent. 

Ans,  $904,166. 

20.  $1240.60  for  12  yrs.   11  mo,  29  da.,  at  7V  per  cent. 

Ans.  $2449.926. 

Case  6.  To  find  the  interest  on  Xotes,  Bonds,  etc.,  on  which  par- 
tial payments  have  been  made. 

Obs.  11.  A  I^OTE  on  ^o^n  is  an  instrument  of  writing ,  in  which 
the  debtor  promises  t^  pay  the  creditor  his  due,  in  such  a  manner,  and 
at  such  a  time  as  may  be  agreed  upon. 

Remark  1.  The  debtor  is  the  one  who  borrows  the  money,  or  in  any  way 
owes  tlie  other;  the  creditor  is  the  one  who  lends  the  money,  or  to  whom  any- 
thing is  due. 


What  id  tt  »otc,  or  bouU?    Wiiich  WiBou  is  tb&Uvbtof?    ^ijicU-  Hm  Gretii* 

VOT? 


Art.  5.  INTEREST.  279 

2.  It  oftan  hnpjus  that  thi  debtor  piys  only  a  part  of  his  uotc,  or  obliga- 
tion, at  a  ti!a3.  Waea  a  payitiMil  is  thu?  anJa  it  is  Cill.^ci  an  enlorseinent,  and 
is  Written  on  ths  back  of  tha  note . 

Obs.  12.  When  the  debtor  pays  the  crelitor  a  part  of  his  note, 
or  obUgation,  it  is  evident  that  the  sun  j^aid,  should  be  deducted 
from  the  sum  das  ;  the  sum  due  hiinr  the  facz  of  the  note,  together 
with  the  inic/'est  of  the  sccniJ  until  the  thm  of  payrmnt. 

Hence,  the  following 

1.    RULE. 

I.  Find  the  amount  of  the  i^rbiclpal  to  the  first  time  when  a  payment 
teas  made,  which  either  alone,  or  together  with  the  preceding  pjayments, 
{if  any,)  exceeds  the  interest  then  due.  • 

ir.  From  this  amount  suJrtra^i  the  payment,  or  sum  of  the  payments 
male  within  the  time  for  which  the  interest  was  computed,  and  the  re- 
mainder will  be  a  new  p>rincipal,  with  which  pro:eed  as  before. 

Note. — The  above,  rule,  withsone  sii-rht  alteration;  of  phrasgoiogy,  is  adopt- 
ed bv  ih'i  Sapretn?  Court  of  the  United  Stales,  and  also  by  the  difTdrent  States  of 
the  Union,  with  but  few  exceptions. 

In  the  followinir  examples,  6  per  cent,  is  understood  unless  other- 
wise stated: 


(O 

$800  Columbus,  Ohio,  Jan.  1st,  1836. 

For  value  received  I  promise  to  pay  Thomas  Trueman,  or  order, 
eight  hundred  dollars,  on  demand,  with  interest, 

Charles  Pay  well. 

On  this  note  were  the  following  endorsements  ; 

Sept.  15th,  18^6,  $300.) 

July  9th,  1837.  820.  V 

Dec.  12th,  1837  ^400.^ 

What  remained  due  Auix.  7th.   1838?  Ans,  $159,655. 


Wh:it  is  an  cndortemf  nt?  Wliero  written?  When  part  of  a  note,  or  ebli- 
p;»tion  iH  paid,  what  fact  iscvid»-nt?  What  is  the  ^um  due?  VVhat  then  i.s  'h© 
rul«  fur  conipuiiiig  iiitercst  ounotes,  bouds,  Ck.c.,whon  pariui  piyuieiitiJUave 
been  niatic? 


2S0 


COMMON    ARITMMETIC. 

Operation. 


Principal 

Time  from  Jan.  1st,  1836  to  Sept.  IStb,  1836,  8 
mo.  14  da,;  (Sect.  IX,  Art.  5,  Obi.  1,)  interest  of 
^800  for  this  time, 


Sect.  XIII 

$8C0. 
33.866. 


Amount,  $833,866. 


Payment,  Sept.  15th,  1836,  exceeds  the  interest, 

Remainder  for  a  ne\y  principal, 

Interest  from  Sept.  16th,  1836  to  July 
9th,  1837,  (9  mo.  24  da.)  826.159, 

Payment,  July  9Lh,  1837,  less  than 
the  interest,  20. 

Interest  from   Sept.  15th,    1836  to  Dec.  12th,  ' 
1837,  (1  yr.  2  mo.  27  da.) 

Amount, 
Payment,  July  9th,  1 837,  $20.  \ 

Payment,  Dec.  12th,  1837,      $400,  \ 

Sum  of  these  payments. 

Remainder  for  a  new  principal. 
Interest  from  Dec,  12th,  1-837  tOx\ug,  7th,  1838, 
(7  mo.  25  da.) 


300 


$533,866, 


Amount  due  Aug,  7th,  1838, 


39.772, 
$j73.638 

420.000 

$153,638 

6.017 

$159,655 


(2.) 


$1200 


Cleveland,  June  29th,  1840. 

For  value  received,  I  promise  to  pay  Timothy  Just,  or  bearer, 
twelve  hundred  dollars,  with  interest, 

James  Goodman. 
Endorsed,  Dec.   18th,  1840,         $100, 
Aug.  1st,  1841,  $250. 

Jan,  7th,  1842,  $10.25.  )■ 

No-/.  3d,  1842,  $600. 

July  1st,  1843, 
What  was  due  Jan,  1st,  1844. 


1.75.  I 


$200.        J 


Ans,  $205,889 


(3.) 


.75,  Cincinnati,  May  2nd,  1843. 

For  Value  received,  six  months  after  date,  I  promise  to  pay  Rob*^ 


Art.  5.  INTEREST.  *  281 

ert  Brown,  or  bearer,  six  hundred  and  twenty -three  dollars,  and 
seventy-live  cents.  John  Smith. 

Endorsed,  Jan.  1st,  1844,  $75.25. 

Dec.  13th,  1844,         8200. 
Aug.  7th,  1-845,  810.75. 

Jan,  1st,  1846,  8350.       J 

And  April  4th  1846,  he  paid  the  balance 
How  much  was  this  balance?  Ans.  870.942. 

Remark  1.  When  a  nottMs  given  without  mantiou  of  interest,  as  tho  above, 
it  is  not  customary  to  charge  intf-rest. 

2.  After  a  note  becomes  due,  however,  if  payment  is  delayed,  it  will  draw 
legal  interest,  although  no  mention  be  made  of  iaturest. 

11.     VERMONT  RULE. 

^^Find  the  amount  of  the  principal  from  the  time  the  note  was  given, 
till  the  time  of  settlement:  next,  find  the  amount  of  each  payment  from 
the  time  the  payment  was  made  till  the  time  of  settlement;  finally,  add  to- 
gether the  amounts  of  the  several  payments  and  subtract  their  sumfrofni 
the  amount  of  the  principal  already  found ;  the  remainder  will  be  the  sum 
due.'' 

(4.) 
81000.  -  Pittsburgh  Feb.  2nd,  1840. 

For  value  received,  I  promise  to  pay  Amos  Bush,  or  bearer,  one 
thousand  dollars,  on  demand,  with  interest. 

Chester  Dale. 

Endorsed,  Nov.  1st,  1840,         8300.^ 

July  5th,  1841,         8275. 

Feb.  19th   1842.       8325. 

Oct.  12th  1842,        8100. 
What  was  due  Jan.  1st.,  18431  893.051. 

Operation. 

Principal,         81000. 
Interest  of  principal  from  Feb.  2nd,  1840  to  Jan. 

1st,  1843;  174.833 


Amount  of  same;  81174,833 


What  is  said  respecting  interest  on  notes  in  which  no  mention  is  made  of 
interest?  What  is  said  respecting  interest  on  notes  which  are  not  paid  when 
they  become  due?  What  is  the  Vermont  rule  for  computing  interest  on 
notcf?  \..is>i5«i  'Ji!'  TK^^s>■:3 1^%  muT.  o 


282 


COMMON    ARrrHMETIC 


Sect.  XIII 


First  payment,  made  Nov.  1st,  1840.     8300,000 

Interest  of  same  till  Jan.  1st,  1843.  39,000 

Second  payment  made  July  5th,  1841.  275.000 

Interest  of  same  till  Jan.  1st,  1843.  24.566 

Third  payment,  made  Feb.  19th,  1842.  325.000 

Interest  of  same  till  Jan.  1st,  1843.  16,900 

Fourth  payment  made  Oct.  12,  1842.  100.000 

Interest  of  same  till  Jan*  1st.  1843.  1.316 


Total  amount  of  payments,  and  interest; 


Amount  due; 


^1081-782 
$93,051 


(5.) 


^710.40.  Burlington.  July  4th,  1841. 

For  value  received,  on  demand  I  promise  to  pay  Edwin  Farr,  or 
order,  seven  hundred  and  ten  dollars,  and  forty  cents,  with  interest. 

George  Hunter. 


Endorsed,  Jan.  1st,  1842, 
March  3d,  1843, 
Feb.   19th,  1844, 

The  balance,  July  4th,  1844. 

Required  the  balance? 


$400. 

$250.50. 
$100. 


Ans.  $5,241 


(6.) 


Buffalo,  March  1st,  1842. 


$400, 

For  value  received,  on  demand  I  promise   to  pay  Ira  Jones,  or 
bearer,  four  hundred  dollars,  with  interest  at  7  per  cent. 

Llwis  Maker. 
Endorsed,  Jan.  1st,  1844,  $215.87^. 

April  19th,  1845,  $212. 

The  balance,  Sept.  1st,  1845. 
Required  the  balance?  Ans.  $39,115. 


L371.  J 


III.    CONNECTICUT  RULE. 

I.  ''Fin J  the  amount  of  the  principal  to  the  time  of  the  first  payment,  if 
it  be  a  year  or  more  from  the  time  the  interest  commenced,  and  Irom  thit 
».mount  subtract  the  payment. 

II.  The  remainder  will  be  a  new  principal;  find  the  amount  of  this  to  th« 
next  payment,  from  which  subtract  the  payment  as  above.  So  continue  to  do 
from  paymiMit  to  payment,  until  all  are  employed,  provided  a  year  or  more  in- 
tervenes between  each  two  payments. 

III.  But  if  the  time  between  any  two  payments  be  less  than  a  year,  find 
the  amount  of  the  last  principal  for  a  year,  and  of  the  payments   up  to  Iha 


What  is  the  Connecticut  rule  for  computing  intereit  on  notea? 


Art.  5.  .  ■,;  iNTEfiEST.  283 

■ame  time,  and  subtract  th«^  latter  from  the  forsner.  If,  howerer,  a  year  over- 
runs the  time  of  settlement,  tind  the  amounts  up  to  ihui  time  instead  of  for  a 
year. 

IV.  If  any  remainder,  after  subtraction,  be  greater  than  th"  preceding 
principal,  the  preceuing  principal  is  still  to  be  continued  ks  the  principal  far  the 
Bucceding  time,  instead  of  the  remainder;  and  the  difference  is  to  be  regarded 
as  so  much  unpaid  interest  which  is  to  be  added  to  the  principal  at  the  titne  of 
the  next  payment." 

(7.) 
$1200.  Hartford,  Oct  7th.  1836. 

For  value  received,  on  demand  I  promise  to  pay  Henry  Inman, 
or  bearer,  twelve  hundred  dollars,  with  interest. 

X  2T£R    RuSS 

Endorsed,  April  3d,  183^,         8600. 

Jan,  1st,  1839,  $300. 

Sept.  2d,  1839,         $250. 

What  was  due  Jan.  12th,  1840?  Ans.  $210,309, 

Operation. 

Principal,  $1200. 

Interest  of  the  same  to  April  3d,  1838,  (17  mo. 

26  da.)  $107.20. 


Amount,  $1307.20. 

1st  payment,  deduct,  600. 


Remainder  for  a  new  principal,  $707.20 

Interest  April  3d,    1839,    (12  mo.)   payment 
being  made  before  a  year  has  elapsed,  42.432 


Amount,  $749,632 

2nd  payment  made  Jan.  1st,  1839,  $300      ^ 
Interest  of  same  to  April  3d,  1839,  v 

(3  mo.  2  da.)  4.60) 

Amount  of  payment,  deduct,  304.60 


Remainder  for  a  new  principal,  $445,032 

Interest  from  April  3d,  1839  to  Jan.  12Lh, 
1840,  (9  mo.  9  da.)  settlement  being  made  within 
the  year,  20.693 


Amount,  $465,725 

3d  payment  made  Sept.  2d,  1839,  $250         ^ 
Interest  of  same    to   Jan.    12th,  V 

1840,  (4  mo.  10  da.)  5.416  S 

Amount  of  payment,   deduct,  255.416 


Remainder  due  Jan  12th,  1840,  :mi^j     $210,309 


284  CO.VIMON    ARITHMETIC.  Sect.    XIII 

(8.) 

S900.  Few  Haven,  May  3d,  1842. 

For  value  received,  on  demand,  I  promise  to  pay  Silas  Weeks,  or   :■ 
bearer,  nine  hundred  dollars,  with  interest. 


John  Leak. 

Endorsed.     June  1st,  1843, 

$300. 

^ 

Feb.  2nd,  1844, 

^250. 

Jan.  1st,  1845, 

$300. 

>• 

Nov.  9th,  1846, 

$100. 

The  balance  Jan.  1st.  1847. 

Required  the  balance. 

* 

Ans.  $76,422. 

By  each  of  the  preceding  rules  let  the  learner  find   the  balance 
due  on  the  following  notes: 

(9.) 

$1500.  Sandusky,  June  1st,  1830. 

For  value  received,  on  demand,   I  promise  to  pay  Jonas  Trusty, 
or  bearer,  fifteen  hundred  dollars,  with  interest. 

Theodore  Hand. 
Endorsed.  Jan.  1st,  1831,         $500.    ) 
July  4th,  1832,        $700.    V 
Feb.  2d,  1833,         $30(y.    ) 
What  remained  due  April  1st,  1833? 

C  By  the  Mass.  and  N.  Y.  Rule,  $164,878.  tsim  -aud 

Ans.  <  By  the  Vermont  Rule,  $153.40. 

(  By  the  Conn.  Rule,  $165,065. 

(10.) 

$12500.  Louisville,  Sept.   1st,  1841. 

For  value   received,    I   promise  to  pay  Henry  Loan,  or  bearer, 
twelve  thousand,  five  hundred  dollars,  with  interest. 

Joseph  Fairman. 
Endorsed.  Jan.  1st,  1842,     $1800.1 
Oct.  9th,  1843,     $6000.  | 
Dec.  15th,  1843,  $     50.  J-     . 
Nov.  25th,  1844,  $3000.  I 
Sept.  30th,  1845,  $1500.  J 

What  remained  due  Jan.  1st,  1846? 

C  By  the  Mass.  and  lY.  Y.  Rule,  $2187.571. 

Ans.  <  By  the  Vermont  Rule,  $1939.1 18. 

(  By  the  Conn.       >    -    - ,  Rule,  $2202.525. 


Art.  5.  iNTEUEST  285 

Case  7.     Pro^jlems  in  Interest. 

Obs.  I.?*  A  Problem  ^5  a  question  proposed  requiring  a  solu- 
tion^ 

This  question  may  apply  either  to  the  investigation  of  some  truth, 
or  principle,  or  to  the  operation  of  practical  questions* 

RexViark. — In  the  preceding  example*  we  notice  five  different  terms,  or  parts 
Cinnected  with  Interest;  viz:  the  Principal,  \.\\q  Time,  iht  Rate,  the  Interest, 
and  the  Amount.  The  relation  of  these  terms  to  each  other,  is  such,  that  if  any 
three  are  given,  the  other  two  can  be  found. 

Problem  !♦ 

The  Principal,  Bate  per  cent,  and  time  being  given,  to  find  the  Inter- 
est, and  Amount. 

This  is  deemed  the  most  important  problem.  It  has  already  been 
exemphfied  in  the  preceding  Cases  of  this  Article. 

Problem  II. 

The  Principal,  Interest,  and  Time  being  given,  to  find  the  rate  per 
cent. 

K  A  man  borrowed  8250  for  3  years,  and  paid  845  interest. — 
What  was  the  rate  per  cent.?  Ans*  6. 

Solution.— "IhQ  interest  of  $250  for  3  years  at  1  percent  is  87.50. 
(Obs*  10.)  Therefore,  as  many  times  as  87.50  is  contained  in  845, 
so  many  times  more  than  1  per  cent,  was  the  per  cent,  paid  ;  and 
845-r-87.50  =  6. 

Proof.— 3  yrs.=  36  mo.  36-^2  =  80.18;  8250X*18  =  845 
interest.     (General  Rule,  Case  4.)     Hence — 

To  find  the  rate  per  cent»  in  such  cases. 

Obs.  1 4.  Divide  the  given  interest  by  the  interest  of  the  given  prin- 
cipal, for  the  given  time,  at  1  per  cent. 

Remark. — The  amount  cb^w  be  foundry  nddintr  together  the  principal  and  in- 
terest; {()hs.  7,  Remark)  also,  the  principal  can  always  be  found  by  subtracting 
the  interest  from  the  amount  and  the  interest  can  always  be  found  by  subtracting 
the  principal  from  the  an-.o  int.     (Sect.  VI,  kn.  1,  Obs.  11.) 

2.  If  I  borrow  8420  for  2  yrs.  6  mo.,  and  pay  873.50  interest, 
what  is  the  rate  per  cent.?  Ans.  7. 

What  is  a  problem?  To  what  may  this  question  apply?  How  many  terms 
are  connected  witfi  intpr»-pt?  Name  them.  Whtit  relation  hfive  these  terms 
to  each  other?  Which  is  the  most  importnnt  problem?  Why  do  vou  suppose 
it  to  be  the  mo-!t  important?  How  do  we  find  the  rate  per  cent,  when  the 
principal,  iutert>st,  and  tirnf»  arii<  ^ivpn?  How  can  the  amount  be  found  in 
such  casts?    The  principal?    The  interest?     awnentii «^i  s^i^ut  u    i«w*tT|^©*s 


:286  COMMON    ARITHMETIC.  ScCt.    XIII 

3.  If  $117.60  is  pall  for  the  use  of  S700  3  yvs.  8  mo,  24  da., 
what  is  the  rate  per  cent.?  x\ns,   4^-. 

4.  If  a  person  pays  8151.20  for  the  use  of  $810  2  yrs.  4  mo., 
what  rate  per  cent  does  lie  pay?  Ans.   8. 

-    5.   A  man  lent  $1000,  and  at  th.e  end  of  3  yrs.  6    mo.    received 
for  his  due  $1230.25.     Required  the  rate  per  cent.? 

Ans.  6|. 

6.  Lent  $750  for  6  mo.,  and  receive]  for  principal  and  interest 
$765.     Required  the  per  cent.?  Ans.   4. 

7.  At  what  rate  per  cent,  must  81250  be  loaned,  in  order  to 
amount  to  ^1343.75  in  1  yr.  6  mo.?  Ans»  5. 

Problem  III. 

The  Principal,  Rate  per  cent.,  and  Interest  being  given  to  find  the 
Time. 

1.  A  person  borrowed  ^400  for  a  certain  time  at  7  per  cent,  in- 
terest: at  the  end  of  the  time  the  interest  was  ^70.  How  long  did 
he  have  the  money?  Ans.  2  yrs.  6  mo. 

Solution. — The  interest  of  $400  for  1  year,  at  7  per  cent., is  ^28, 
(Obs.  7.)  Then  if  the  interest  of  ?t^400  for  1  year  is  $28,  it  will 
take  it  870 -^  $28  =  2^  yrs  to  produce  $70  interest. 

Proof.— 2^  yrs.  =  30  mo.  30-^-2  =  $0.15;  $400  X  -15=  $60 
interest  at  6  per  cent.  (Obs.  8.)  $60 -f- 6  =  $10  interest  at  1  per 
cent.     $10  X  V  =  '*^70  interest  at  7  per  cent.    (Obs.  10.)  Hence— 

To  find  the  time  in  such  cases: 

Obs.  15.  Divide  the  given  interest  by  the  interest  of  the  principal 
for  1  year  at  the  given  rate. 

Remark. — If  there  i.s  a  remainder  after  dividing,  it  may  either  be  considered 
as  the  fractional  part  of  a  year,  or  continued  to  decimals,  fn  either  case  the 
valuecan  easily  be  found  in  months  and  days.  (Sect.  IX,  Art.  3,  Case  2,  Ex. 
7,  and  Case  3.     Rule.) 

2.  In  what  time  will  $375  gain  $90  interest,  at  6  per  cent.?      '■-'"' 

Ans.  4  yrs. 

3.  In  what  time  will  $700  gain  $210  interest  at  9  per  cent? 

Ans.  3  yrs.  4  mo. 

4.  In  what  time  will  $560  gain  $45.36  interest,  at  4^-  per  cent.? 

Ans.   1  yr.  9  mo.   18  da. 

5.  In  what  time  will  $840  amount  to  $10 15.08  J  at  5 J  per  cent.? 

Ans.  3  yrs.  7  mo.  15  da. 


How  do  we  find  the  time,  when  the  principal,    rate  per  cent.,  and  interes 
ere  given?    If  there  is  a  remainder  after  divkiing,  what  is  done  with  it?         j 


Art.  5.  INTEREST.  287 

6.  In  what  tirae  will  f-OCO  amount  to  $1710,  at  10  per  cent.? 

Alls.  9  yrs. 

7.  In  what  time  will  $1200  amount  to  $2200,  at  8i  per  cent.? 

Ans.    10  yrs. 

8.  In  what  tirae  will  any  sum  of  money  double  itself,  at  5  per 
cent.? 

Suggestion. — Assume  SlOO  as  the  given  sum.  Then,  in  what 
tirae  will  $100  gain  $100  interest,  at  6  per  cent.? 

Ans.  20  yrs. 

9.  In  what  time  will  any  sum  of  money  double  itself,  at  8^  per 
cent.?  Ans.   12  years. 

10.  In  what  time  will  any  sum  double  itself,  at  6  per  cent.? 

Ans.  16  yrs.  8  mo, 

11.  In  what  time  will  a;  y  sum  double  itself,  ;  tthe  following  rates 
per  cent:  1;  4;  7;  3;  9;  2;  8;  10;  12;  15;  20;  24;  25;  30;  40; 
and  50  per  cent.?  At  \\;  2^;  3^;  4|;  b\;  1\;  8^;  and  12^  per 
cent.? 

Ans.  in  order.  100  ys.;  25  yrs.;  Hf  yrs,  33  yrs.  4  mo.;  \\\ 
yrs.;  50  yrs.;  12  yrs.  6  mo.;  10  yrs;  8  yrs.  4  mo.;  6  yrs.  8  mo.;  5 
yrs.;  4  yrs.  2  mo.;  4  yrs.;  3  yrs.  4  mo.;  2  yrs.  6  mo.;  2  yrs.; 
66  yrs.  8  mo.;  40  yrs.;  28^  yrs.  22^  yrs.;  18/y  yrs.;  13  yrs.  4  mo.; 
lljY  yrs.;  8  yrs. 

12.  In  what  time  will  any  sum  treble,  aud  quadruple  itself,  at  12^ 
per  cent. 

^   ,    ^Treble  itself  in  16  yrs. 

\  Quadruple  itself  in  24  years. 

Problem  IY. 

The  Interest,  Time,  and  Rateper  cent,  being  given  to  find  the  Prin- 
cipal. 

1.  The  interest  of  n  certain  note  is  $30,  it  having  been  at  inter- 
est 2  yrs.  at  5  per  cent.  Required  the  face  of  the  note,  or  princi- 
pal? 

Ans.  $300. 

Solution. — The  interest  of  $1  for  two  years,  at  5  per  cent,  is 
$0.10.  (Obs.  7.)  Then  the  principalis  as  many  times  $1,  as  $0.10 
is  contained  in  $30;  and  $30  -f-.  10=  $300.     Hence— 

To  find  the  principal  in  such  cases: 

Obs.  16.  Divide  the  given  interest  hy  the  interest  cf  %\  for  the 
given  time,  and  at  the  given  rate  per  cent. 

How  do  we  fiud  the  principal,  when  the  interest,  time  aud  rate  p^r  cent,  are 


288  COMMON    ARITHMETIC.  ScCt.    XIII 

2.  What  principal  at  6  per  cent.  Avill  gain  ^27.45  in  1   yr.  6  mo.? 

Ans.  S305. 

3.  What  principal  at  4\  per  cent,  will  gain  1^76.50  in  3  yrs.  4  rao. 
24  da.?  Ans.  ^500. 

4.  What  principal  at  7  per  cent  will  gain  $236.25  in  4  yrs.  6  mo.? 

Ans.' $750. 

5.  What  principal  at  9  per  cent   will  gain    $237   in  2  yrs.  7  mo. 
18  da.?  Ans.  $1000. 

6.  What  principal  at  12^  per  cent  will  gain  $1265.625  in  6  yrs- 
9  mo.?  Ans.  $1500. 

Problem  V  .^ 

The  Amount,  Rate  per  cent.,  and  Time  hsing  given  to  find  the  Prin- 
cipal and  interest. 

The  amount  of  a  certain  note  is  $660  ;  it  has  been  on  interest 
1  yr.  8  mo.  at  6  per  cent.  Required  the  face  of  the  note,  and  the 
interest?  Ans.  Face  of  the  note  $600;  and  the  interest  $60. 

Solution. — The  amount  of  $1  for  1  yr.  8  mo.  at  6  per  cent.,  is 
$1.10.  Then  the  principal,  or  face  of  the  note,  must  be  as  many 
times  $1,  as  $1.10  is  contained  in  $660;  and  $660 -^$  1.10  =  $600 
the  principal,  and  $660 — $600=  $60  the  interest.     Hence — 

To  find  the  principal  and  interest  in  such  cases  : 

Obs.  17.  Divide  the  given  amount,  by  the  amount  of  $1  for  the 
given  time,  at  the  given  rate  per  cent. 

To  find  the  interest ;  Subtract  the  principal fromthe  amount.  (Obs. 
14.  Rem.) 

2.  What  sum  of  money  will  amount  to  $77.76,  in  1  yr.  4  mo.  at 
6  per  cent?  -  Ans.  72. 

3.  What  sum  of  money  will  amount  to  $1 190  in  5  yrs.  at  8  per 
cent?  Ans.  $850. 

4.  The  amount  of  a  certain  note  is  $776.3?;  it  has  been  on  inte:N 
est  4  yrs.  8  mo.  24  da.  at  4^  per  cent.  Required  the  face  of  the 
note  and  the  interest? 

Ans.  Face  of  the  note  $640.     Interest  $136.32. 

5.  The  amount  of  a  note,  after  having  been  on  interest  3  yrs.  10 
mo.  H  da.  at  7  per  ceni  was  $1144.65.  Required  the  face  of  the 
note,  and  the  interest. 

Ans.  Face  of  the  note  f?990.     Interest  $244.65. 

How  do  we  find  the  principal,  when  the  anfiouiit,  rate  per  cent  and  time  are 
given?    Haw  do  we  find  the  interest? 


Art.    5.  INTEREST.  SBS 

6.  If  the  amount  of  a  note  is  S2139  after  having  been  on  interest 
7  yrs.  9  mo.  27  da.  at  10  per  cent.,  what  is  the  face  of  the  note, 
and  what  the  interest?  • 

Ans.  Face  of  the  note  ^1200.     Interest  393P. 

DISCOUNT. 

Obs.  18.  Discount  is  a  deduction  made  from  a  debt  for 'paying  it 
before  it  becomes  due. 

Thus,  if  I  owe  a  man  $106,  due  1  year  hence,  and  he  wishes 
me  to  pay  him  now,  it  is  evident  he  ought  to  make  some  deduction 
for  present  payment.  And  I  ought  only  to  pay  him  such  a  sum  as 
would  amount  to  ^106  in  l-  year.  If  the  discount  is  made  at  6  per 
cent.,  for  instance,  I  ought  to  pay  him  $100,  because  the  interest  of 
^100  for  1  year  is  86,  and  the  amount  is  $100+ 86  =$106. 

Obs.  19.  The  sum,  which  in  the  given  time,  with  its  inter3st, 
would  amount  to  the  sum  on  which  the  discount  is  made,  is  cal-^d 
the  Present  Worth.  Thus,  in  the  example  mentioned  above,  $106 
is  the  sum  on  which  the  discount  is  made,  $6  is  the  discount,  and 
$  100  the  present  worth. 

1.  What  is  the  present  w^orth  of  $440,  payable  in  1  yr.  8  mo., 
discounting  at  6  per  cent.     What  is  the  discount? 

Ans.  Present  Worth  $400.  Discount  $40. 
Suggestion. — By  examining  this  question  attentively,  we  perceive 
that  it  does  not  differ  from  those  under  Problem  Y;  i\\Q  Debt  corres- 
ponds to  the  Amount;  the  present  worth  to  the  Principal;  and  the 
Discount  to  the  Interest  for  the  given  time,  at  the  given  rate  per 
cent.  Therefore,  $440^  $1.10  =  $400  the  present  worth;  and 
^440~-.$400  =  $40  the  discount. 

Remark. — It  is  not  unfrequently  supposed  thatif  we  find  the  interest  on  tha 
given  sum  for  the  fjiven  time,  that  this  interest  will  be  the  discount,  which  sub- 
tracted from  the  given  sum,  will  give  the  present  worth.  But  this  error  may  be 
avoided  if  the  pupil  will  recollect  that  the  debt  answers  to  the  anwunt'xn  inter- 
est, and  the  principal  is  altmys  the  sum  on  which  the  interest  is  calculated. 

2.  What  is  the  present  worth  of  $I'6Q  payable  in  10  mo.  at  6  per 
cent.?  Ans.  $952,38^.  ;t- 

3.  What  is  the  present  worth  of  $1488,  due  4  years  hence?      ^  ^ 

Ans.  $1200. 
In  these  examples,  6  per  cent,  is  understood  when  no  other  rate 
is  mention  d. 

What  is  discount?     What  is  the   present  worth?     In  the  example  given 

which  is  the  sum  on  which  the  discount  is   made?     Which  is  the   discount? 

Which  is  the  present  worth?    What  error  do  many  fall  into  respecting  discount? 

How  may  'ti^lii  error  be  avtHded  T  ,, 

14 


290  COMMON    ARITHMETIC.  Scct.    XIII 

4.  What  is  the  present  worth  of  ^1590  due  1  yr.  hence? 

Ans,  $1500, 

5.  What  is  the  present  worth  of  ^1360  due  6  yrs.  hencc;? 

Ans.  i$1000. 

6.  What  is  the  discount  of  $872  due  2  yrs.  hence? 

Ans.  893.429.+ 

7.  Whaat  is  the  discount  of  'f  1736  due  3  yrs,  6  mo.  hence? 

Ans.  $301,29.+ 

8.  What  is  the  discount  of  $2412.75  due  6  yrs,  1  mo.  24  da. 
hence.?  Ans.  $694,263  + 

9.  A  merchant  bought  goods  to  the  amount  of  $1968.75,  pay- 
able in  10  mo.     How  much  ready  money  should  pay  the  debt? 

Ans,  $1875. 

10.  What  is  the  present  worth  of  $2303.40  due  2  yrs,  4  mo. 
hence,  discount  being  made  at  7  per  cent?     The  discount? 

Ans.  Present  worth  $1980.     Discount  $323.40. 

11.  What  is  the  present  worth  of  $2156.37^  due  3  yrs.  7  mo.  18 
da.  hence,  the  money  being  wor^h  9  per  cent.?     The  discount? 

Ans.  Present  worth  $1625.     Discount  $531 .3  7. 

12.  A  man  sold  a  farm  for  $5836.80,  payable  in  3  yrs.  6  mo.; 
he  could  have  received  $4650  down  for  it.  Did  he  gain  or  lose  by 
selling  on  credit;  and  how  much,  the  money  being  worth  8  per  cent.? 

Ans.  He  lost  $90. 

13.  What  is  the  difference  between  the  interest  of  $2352  for  4  yrs. 
8  mo.  12  da.,  at  10  per  cent.,  and  the  discount  of  the  same  sum,  for 
the  same  time,  and  at  the  same  rat^?  Ans.  $353.44. 

14.  A  merchant  bought  a  lot  of  goods  for  $2500  ready  money, 
and  sold  them  the  same  day  for  $3044.25  payable  in  1  yr.  9  mo.  12 
da,  at  6  per  cent  interest.  Did  he  gain  or  lose  by  the  transaction, 
and  how  much?  Ans.  He  gained  $250. 

When  payments  are  to  be  made  at  different  times,  to  find  the 
present  worth: 

Obs.  20.     Find  the  present  worth  of  each  payment  separately,  an 
add  together  these  results,  for  the  present  value  of  the  debt. 

This  rule  is  so  evident  that  it  needs  no  demonstration,  Let  tliQ 
learner  see  if  he  cannot  demonstrate  it  himself. 

16.  What  is  the  present  worth  of  $3000,  one  half  payable  in  2 
yrs.  6  mo.  18  da.,  and  the  other  half  payable  in  3  yrs.  4  mo.  24  da.? 
Tie  discount? 

Ans.  Present  worth.  $2546.801+.     Discount  $453,199+. 

16.  A  merchant  bought  goods  to  the  amount  of  $2460;  \  of 

When  payments  are  made  at  different  times,  how  do  we  find  the  present 
worth? 


Art    5.  INTEREST.  291 

Avhieh  wa>  to  be  paid  in  4  mo. ;  j  in  8  mo.;  and  |-  in  12  mo.     What 
sum  sliouivi  be  paid  at  the  present  time  to  balance  tiie  debt? 

Aiis.   ^2365.966-(-. 

17.  An  agent  has  S12G0  to  use  m  trade,  and  is  to  receive  5  per 
cent,  on  what  he  lays  out.  What  sum  does  he  lay  out,  and  how 
much  does  he  receive? 

Ans.  He  lays  out  $  1 200,  and  receives  §60. 

Solution. — As  often  as  he  lays  out  ^1,  he  receives  ^0.05.  There- 
fore, $1260-J- ^1.05  ==  ^1200  what  he  lays  out,  and  81260— 
^  1 200  =  $60  what  he  receives.  Had  we  multiplied  8 1 260  by  .05, 
(as  many  a  c  rcless  pupil  would  have  done,)  we  should  have  given 
the  agent  5  cents  for  every  95  that  he  laid  out,  instead  of  100  as 
required  by  the  quest  on.) 

Proof.— $1200  X -05  =  $60.     $1200  +  $60  =  $1260 

18.  An  agent  has  $918  to  use  in  trade,  for  which  he  receives  8 
per  cent  on  what  he  lays  out.  How  much  does  he  lay  out  for  his 
employer,  and  how  much  does  he  receive? 

Ans.  He  lays  out  $850,  and  receives  $68. 

Case  8. — Banking. 

Obs.  21.  A  ljky.K  is  an  institution  that  deals  in  money .  lis  Cap- 
ital or  Sto'k  is  divided  mto  ])aYts  c^iWad  Shares,  and  owned  by  in- 
dividuals cabled  Stockholders.     A  share  is  usually  $100. 

Obs.  22.  The  operations  of  ;.  bank  are  conducted  by  a  President 
and  JBoanl  of  Directors.  It  has  a  deposite  of  ;:pecie,  and  issues 
notes,  or  bills,  intended  to  be  used  as  a  circulating  medium  instead 
of  gold  and  silver.  These  bills  are  usually  obtained  from  the  bank 
in  loans. 

Obs.  23.  In  loaning  money,  the  banks  usually  deduct  the  legal 
irderest  off  the  face  of  the  note  for  the  given  time,  and  pay  the  hold- 
er the  balance.  The  sum  which  the  holder  of  the  note  receives,  is 
called  the  Proceeds  of  the  note.  The  sum  deducted  is  called  the 
Bank  Discount,    and  is  the  same  as  simple  interest  in  advance. 

Obs.  24.  The  time  from  the  date  of  a  note  until  it  becomes  due, 
is  called  the  Days  to  run.  Besides  the  time  mentioned  in  the  note, 
three  additional  days  are  allowed  before  it  is  legally  due.  These 
are  called  Days  of  Grace'^;  and  as  notes  are  not  usually  pad  until 
the  last  of  these  dates,  the  banks  charge  interest   on   the  days  of 


Wlial  is  a  bank?  llow  is  the  capital  divided?  What  are  stockholders? 
How  much  is  a  share?  How  are  the  operations  of  a  bank  conducted?  What 
has  it?  What  does  it  issue?  For  what?  How  are  these  bills  usually  obtained? 
What  is  usually  done  by  the  banks  in  loaning  money?  What  are  the  pro- 
ceeds of  the   note?     The   bank   discount?     Days  to  run?     Days   of   grace? 

•Sofii9;Bank8  allo\v  4  days  of  Grace. 


292  COMMON    ARITHMETIC.  Sect.  XIII 

Obs.  25.  A  note  is  said  t-)  be  discounted  when  the  Bank  Dis- 
count has  been  deducted.  The  difference  between  the  Bank  Dis- 
couni  and  ihe  true  discoimt,  is  the  interest  of  the  true  discount  for 
the  given  time.  This  gives  an  extra  pioiit  to  the  Bank.  On  small 
sums  for  a  short  time,  however,  the  diflt^rence  is  trifling  ;  but  when 
the  sum  is  large,  and  the  time  for  whicii  it  is  discounted,  is  long, 
the  ditierence  amounts  to  a  considerable  sum. 

1.  What  is  the  bank  discount  on  a  note  of  $200  payable  in  1  yr. 
6  mo.  at  6  per  cent.?     What  are  the  proceeds? 

Operation. 

The  interest  of  $200  for  1  yr.  6  mo.  is  $18.00 

.10 


The       " 

'*  3  days  of  o-race  is 

Bank  discount, 

Face  of  the  note     $200.00 
Bank  discount             18.10 

$18.10 


Proceeds,  $181.90 

Hence — To  find  the  bank  discount  of  any  note,  draft  <fec.: 

Obs.  26.     Add  the  days  of  grace  to  the  time  the  note  has  to  run 
and  find  the  interest  of  the  face  of  the  note  for  this  time. 

To  find  the  proceeds — 

Deduct  the  Bank  discount  from  the  face  of  the  note. 

2.  What  is  the  bank  discount  of  a  note  of  $500,  payable  6  mo. 
afterdate?  Ans!  $15.25. 

Note. — In  these  questions,  3  days  of  grace,  and  6  per  cent,  are  understood, 
unless  otherwise  stated. 

3.  What  is  the  bank  discount  on  a  note  of  $350  payable  in  2  yrs. 
4  mo.]  Ans.  $49,175. 

4.  What  are  the  proceeds  of  a  note  of  $400,  payable  9  mo,  after 
date?  Ans.  $381.80. 

5.  What  are  the  proceeds  of  a  note  of  $750  for  1  yr,  8  mo.  21 
da.?  Ans.  $672. 

6.  What  is  the  bank  discount  on  a  note  of  $900,  payable  30  days 
after  date,  at  7  per  cent?     The  proceeds? 

.         i  Discount  $5,775. 
^^^'  I  Proceeds  $894,225. 

7.  What  is  the  bank  discount  on  a   note  of  $1500,  payable    90 

When  is  a  note  said  to  be  discounted?  What  is  the  difference  between  the 
bank  discount,  and  the  true  discount  of  a  note?  How  do  we  find  the  bank 
discouut  of  a  note,  draft,  &c.?     The  proceeds? 


Art.    5.  INTEREST.  293 


A         ,  Discount  $19,58i. 
^^^*  ^ 'Proceeds  8I480.41|. 


days  after  date,  at  5  per  cent,  and  allowing  4  days  of  grace?     The 
proceeds? 

8.  What  is  the  diflference  b'rtween  the  bank  discount  of  <*§1200  for 
3  yrs.  6  mo.,  and  the  true  discount  of  the  same  sum  for  the  same 
time?  Ans.  $44,335. 

9.  What  is  the  bank  discount  of  $1000  payable  6  months  aft^r 
date?  Ans.  $30. 50. 

10.  What  are  the  proceeds  of  81260.50,  payable  90  days  after 
date?  Ans,  $1240.963. 

1 1.  A  man  bought  1250  barrels  of  flour  at  $4.50  a  barrel,  cash, 
and  sold  them  the  same  day  for  $5  a  barrel,  payable  in  9  months, 
without  interest.  He  got  his  note  discounted  at  the  bank,  wh;  n 
money  was  worth  6  per  cent.  Did  he  gain  or  lose  by  the  operation, 
and  how  much] 

Ans.  He  gained  $340.62^-. 

12.  A  man  got  a  note  of  81500  payable  in  90  days,  without  in- 
terest, discounted  at  the  bank  at  6  per  cent.,  and  put  the  proceeds  at 
interest  for  1  year  at  7  per  cent.  He  renewed  his  note  at  the  bank 
each  90  days,  by  paying  the  bank  discount  at  each  rene-wal.  At  the 
end  of  the  year,  he  received  his  due  on  what  he  had  lent,  and  paid 
his  note  at  the  bank.  Did  he  gain  or  lose  by  the  operation,  and 
how  much?  Ans.  He  gained  $6.71. 

In  this  example  the  money  is  supposed  to  be  worth  7  per  cent,, 
although  he  paid  but  6  per  cent,  discount  at  the  bank. 

Case  9.     Equation  of  Payments. 

Obs.  27,  Equation  of  Payments  u  the  method  of  finding  tJie 
time,  when  two  or  more  payments,  due  at  different  times,  may  le  made 
at  once,  without  injury  to  either  party.  This  time  is  usually  called 
the  mean,  or  equated  time. 

1.  A.  owes  B.  $60,  to  be  paid  as  follows;  $10  in  1  month;  $20 
in  2  months,  and  $30  in  3  months.  He  wishes  to  pay  it  all  at  once, 
in  what  time  ought  he  to  pay  it?  Ans.  2  mo.  10  da. 

Solution. — The  interest  of  $100  for  1  year  is  $6;  for  2  years,  the 
interest  is  $12,  which  is  the  same  as  the  interest  of  $200  for  1  year. 
Also  the  interest  of  $100  for  3  yrs.  is  $18,  w-hich  is  the  same  as  the 
interest  of  $300  for  1  year.     Hence,  universally — 

Ob-.  28.      The  interest  cf  any  sum  of  money  for  any  time,  is  equal 

to  the  littered  oftioice  os  rnudujor  I i at f  litis  time;  thrte  times  as  much, 

Wh^t  is  equatiou  of  payments?     What  is  the  timo  usually  called? 


294  COMMON    ARITHMETIC.  Scct.    Xlll 

for  onetJi'rdof  tJds  time,  d-c.  And  conversely  ;  The  interest  of  an?/ 
svm  of  money  for  any  time  is  equal  to  the  interest  of  one-haf  as  much 

for  twice  this  time  ;  one-third  as  w.iich  for  three  times  this  time,  d-c. 
Therefore, 

The  int,  of  810  for  I  mo.,  is  the  sarae  as  the  int.  of  81  for  10  mo. 
The       *'      f  20  '*  2         "  "  ''      '^1   for  40  mo. 

The        ''      830   "  3  *'  ''  '^      5$1   for  90  mo. 

And  the  int,  of  860  for  tlic  mean  time,  i^  the  same  as  t])e 

int.  of  $1  fo-  140  mo. 

But  8G0  is  60  times  $1,  therefore,  it  will  take  it  but  ^'^  jiiirt  of  this 
time  to  gain  the  same  interest,  and  gV  of  140  mo.  is  2  mo.  10  da. 

Proof.— The  interest  of  860  for  2  mo.  10  da.  is  §60  X  01 15.  = 
80.70.  (Case  4,  Rule.)  The  interest  of  ^\  for  140  mo.  is  ^"l  X 
.70  ==  f.0.70.     (Case  4,  Rule).     Hence— 

To  find  the  mean  or  equated  time  of  several  payments: 

Obs.  29.  Multiply  each  -payment  by  the  time  before  it  becomes  dae^ 
and  divide  the  sum  of  the  products  by  the  sum  of  the  payments. 

Remark  1.  This  rule  is  founded  on  the  .supposition  that  the  interest  of  any 
sum  of  money  for  a  certain  time,  is  equal  to  the  discount  of  the  same  sum 
for  the  same  time,  Tliis  is  not  the  case,  as  the  discount  is  always  less  than  the 
interest;  the  difference  beinor  equ:^!  to  tiie  interest  of  t'>.^  tru-:^  discount  for  the 
given  time.  The  dilFerence  in  small  sumis,  for  a  short  tiai.',  however,  is  too 
slight  to  be  noticed. 

2.  A  merchant  has  owing  to  him  Si  000,  to  be  pit  1  as  follows: 
$200  in  3  months  ;  ^300  in  5  months;  $250  in  6  months,  and  $250 
in  8  months.  It  is  agreed  lo  make  but  one  payment,  In  what 
time  must  this  payment  be  made?  Ans.  5  mo.  18  da. 

3.  A.  owes  B.  $120  to  be  paid  in  5  months  ;  S84  to  be  paid  in  8 
months;  $132  in  9  montlis  ;  and  $160  in  10  months.  Required 
the  mean  time  for  the  payment  of  the  whole? 

Ans.   8  mo.   5  da.-f- 

4.  A.  owes  C  $1200  to  be  paid  as  follows  :  ^  in  6  months  ;  ^  m 
8  mo.,  and  the  balance  in  10  mo.  Required  the  equated  time  for 
the  payment  of  the  whole. 

Ans.   7  mo.  10  da. 

5.  A  merchant  in  Columbus,  Ohio,  orders  goods  from  New  York 
as  follows: 


To  what  is  tli»'  interest  of  any  sum  of  money  for  any  time  equal?  FJow^ 
do  we  (ind  the  mean  or  equated  time  of  several  payments?  Upon  what  is  this 
rule  founded?  Is  this  correct?  Why  not?  Why  then  is  the  rule  used? 
What  is  the  correct  rule? 


Art.  5.    -^  .  INTEREST.  295 

Oct.   IStli,  1849,  to  the  value  of  $100,  payable  in  6  mo. 

Oct.  25Lh       "         **  "  150.         "  4  mo. 

Nov.  10th       ''         *'  ".  300,         "  3  mo. 

After  ordering  the  latter  quantity  he  wishes  to  give  his  note  for  the 
entire  lot.  Required  the  date  to  which  the  note  should  run  that 
neither  party  may  lose  by  the  operation. 

Ans.  Feb,  25th,  1850. 

Solution. — He  evidently  owes  the  whole  amount  from  Kov.  10th. 
From  Oct.  12th,  to  Nov,  10th,  is  28  days;  from  Oct  25tli  to  Nov. 
10th  is  14davs.  The  accounts  then  stand  ^100  for  152  da.  (180- 
28)  ;  .$150  for  106  da.  (120-14);  and  ^300  for  90  da.  Theequa- 
ted  time  is  then  found  as  usual.  The  date  is  found  by  the  table  on 
jpage  273. 

As  we  have  before  said,  the  rule  given  is  not  strictly  correct.  The 
followinof  will  cfive  the  correct  mean  time  if  no  mistake  is  made  in 
the  operation. 

Obs*  30.  F  hid  the  present  worths  of  the  several  debts  at  the  given 
rate.  Tiwii  find  in  what  time  at  the  same  raJte  the  sum  of  the  present 
worths  would  amount  to  the  sum  of  the  debts.  The  restUt  vnll  be  the 
equated  time. 

Remark.  When  tha  date  of  the  several  transactions  is  different,  the  date 
from  which  to  make  l!ie  calculations  must  be  found  as  in  Ex.  5.  The  date 
from  which  we  calcuhite  in  this  example  is  Feb.  lOth.  Find  the  answers  to 
the  followinor  questions  by  both  Rules. 

6.  Required  the  date  for  the  payment  of  the  following  sums  at  one 
time. 

$300,  payable  in  8  mo.;  date  Jan.  6th,  1849. 
450,       '      "      6  mo.;     "    Jan.  20th,     " 
500,  *'      4 mo.;     "    Feb.  1st, 

1000,         -   ♦*      3  mo.;     "   Mar.  3d, 
Ans.  By  Obs.  29,     June  24th,   1849.      By  Obs.  30,  June  2oth 
1849. 

7.  Required  the  date  for  the  payment  of  the  following  sums,  at 
one  time. 

8800,  payable  in  1  yr.,-  date  Aug.  12th,  1849. 
1000,      '       "      9  mo,;       ''  Sept.  1st, 
loOO,             "      8  mo.;       *'  Sept.   15th,     '' 
1800,             "      6  mo.;       '*  Oct.   1st., 

Ans.  By  Obs.  29.  May  16th,  1850.  By  Obs.  30,  May  17th, 
1850,  nearly. 

8.  Required  the  date  for  the  payment  of  the  following  stuns  at 
one  |ime. 


296  COMMON   AlUTHMETIC.  Scct.   XIII 

^1000,  payable  in  11  mo.;  date  Jan.  1st,  1849. 


Jan.  15th 
Feb,  1st  *' 
Feb, 22d,  " 
March  1 8th  " 
April  9  th, 


1200,  «  10  mo. 

1500,  "  9  mo.; 

1800,  "  8  mo.; 

2000,  "  6  mo.; 

2500,  '•'  3  mo.; 
Ans.  By  Obs.  29.     Sept.  27th  1849;  by  Obs.  30.     Sept.  28th, 
1849, 

COMPOUND  INTEREST. 
Obs,  31.     CoMPOui^D  Interest  is  the  interest,  on  both  the pri7icipal 
and  interest f  when  the  latter  is  notj^ciid  at  the  time  it  becomes  due. 

It  is  calculated  by  adding  the  interest  to  the  principal  at  the  end 
of  each  year,  or  other  stated  time,  and  making  their  5Mm  the  princi- 
pal for  the  next  succeeding  year  o  •  stated  time. 

The  compound  interest  is  found  by  subtracting  the  Urst  principal 
from  the  last  amount, 

1.  What  is  the  compound  interest  of  ^1^400  for  4  years,  at  6  per 
cent?  Ans.  ^104.99. 

Operation. 
•^400  =  Principal. 
f400X-06=  24  =  Interest  for  1  year.     (Obs,  7.) 

$424  :=::  Amouut  for  1  year. 
^424 X. 06  =  25.44  =  Interest  for  2d  year. 

-li/ 

'^449.44  =  Amount  for  2d  year. 
^449.44  X  .06  =      26.966  =  Interest  for  3d  year. 


$476,406  =  Amount  for  3d  year. 
8476.406  X  .06=    28.584  =  Interest  for  4th  year. 


8504.930  =  Amount  for  4th  year. 
Deduct  8400.0C0  =  First  principal. 


Ans.  8104.99  =  Compound  interest  for  4  yrs. 


?^#- 


2.  What  is  the  compound  interest  of  8300  for  3  yrs.  at  7  percent? 

Ans.  867.5129. 

3.  What  is  the  compound  interest,    and  amount  of  $500  for   4 
yrs,  at  5  per  cent? 

Ans,  Amount.  8607.647.     C,  Int.  8107.647: 

What  is  compound  interest?     How  is  it  calculated?     When  months  and  days 
are  givei.  huw  do  we  proceed? 


Art.  5. 


COMPOirND    INTEREST. 


297 


4.  What   is  the   con-po'ind  interest,   of    8700   for  5  yrs.,   at   6 
percent.?     The  amouiit.'' 

Ans.  Amt.  ^^936. 762.     Comp.  Int.  $236,762, 

When  months  and  days  are  given  : 

Obs.  32.     First  find  tJie  amount  for  the  required  number  of  years, 
then  on  the  last  amount  for  the  months  and  days. 

What  is  the  compound  interest  of  -$800  for  5  yis.  6  mo.  24  da., 
at  6  per  cent.?     The  amount? 

Ans.  Amt.  81106.98.     Compt.  int.  8306.98. 

6.  What  is  the   compound  interest  of  8500  for  3  yrs,  8  mo.  18 
da.,  at  6  per  cent.?     Tlie  amount? 

Ans,  Amt.  §621.114.     Comp.  int.  8121.114, 

7.  What  is  the  interest  of  8200   for  3  yrs.,  compounded  every 
6  months,  at  6  per  cent?  Ans.  838.79. 

Note. — The  learner  will  obsprve  to  add  the  interest  lo  the  principal,  at  the 
ead  of  each  6  months.     (Obs.  30.) 

8.  What  is  the  amount  and  interest  of  8600  for  5  yrs.,  compound- 
ed every  6  months,  at  "6  per  cent? 

Ans.  Amount.  8806.34.     Comp.  Int.  8206.34. 

TABLE 

Showing  the  amount  of  81,  at  5,  6,  and  7  per  cent,  for  any  num- 
ber of  years  from  1  to  30, 


Yrs. 

5  per  cen 

1    81.050C 

2 

1.1025 

3 

1.1676 

4 

1.2156 

5 

1.2762 

6 

1.3400 

7 

1.4071 

8 

1.4774 

9 

1.5513 

10 

1.6288 

11 

1.7103 

12 

i   1.7958 

13  M.8856| 

14   1   1.9799} 

16 

[  2.07891 

per  cent  7  per  cent 


,9i.c6oo:8i 

1.1236 

1.1910 

1.2624    1 

1.33821    1 

1.4185|    1 

1.50361    1 

1.5938 

1.6894 

1.7908 

1.8982 

2.0121 

2.1329 

2.2609 

2.3965 


0700! 
1449! 

22501 

3107J 

40251 

50071 

60571 

718l| 

8384; 

9671 

1048 

2521 

4098 

5785 

7590 


i       I       I 
Yrs.  j  5  per  cent: 6  per  centj?  per  cent 


5 98  COMMOSr    ARITHMETIC.  ScCt.    XllI 

The  amount  of  $2  (whether  at  simple  or  compound  interest,)  is 
twice  as  much  as  the  amount  of  '$^1  for  the  same  time ;  the  amount 
of  ^3  is  three  times  as  much  as  $1  ;  &c.     Hence — 

To  find  the  amount  of  any  sum  by  the  table: 

Obs.  33.  MultljAi/ the  amount  of  ^\  for  tJie  required  iiumher  of 
years,  hij  the  given  sum     To  find  the  interest:  See  Obs.  30. 

9.  What  is  the  amount  of  $2000  for  15  yrs.,  at  5  per  cent.,  com- 
pound interest? 

Solution.  By  the  table,  the  amount  of  81  for  15  yrs.  is  82.0789 
and  82.0789  X  2000  =  84157.80.  Ans.  84157.80. 

10.  What  is  the  amount  of  82500  for  20  yrs.,  at  6  per  cent, 
compound  interest? 

Ans.  88017.75. 

11.  What  is  the  amount  of  81500  for  30  yrs.,  at  7  per  cent, 
compound  interest?  Ans.  811418.30. 

12.  What  is  the  amount  of  850000  for  18  yrs.  G  mo.  12  da.,  at 
6  per  cent.,  compound  interest?     (See  Obs,  31.) 

Ans.  8147281.88. 
13:  What  is  the  amount  of  810000  for  8  yrs.  8  mo.  at  6  per  cent 
compound  interest? 

Ans.  816575.52. 

14.  What  is  the  amount  of  3000  for  40  yrs.,  at  6  percent.,  com- 
pound interest? 

Ans.  830855.842. 

Note. — First  find  the  amount  for  30  yrs.,  and  then  on  this  amount  for  10 
yrs. 

15.  Required  the  amount  of  85000  for  37  yrs.,  at  1  per  cent., 
compound  interest? 

Ans.  861114.5477. 

Remark. — In  several  of  the  precedinjj  examples  the  principal  has  doubled 
itself.  Any  sum  at  5  per  cent.,  compound  interest,  will  double  itself  in  14  yrs 
2  mo.  1.3  da.-,  at  6  per  cent,  in  11  yrs.  10  mo.  22  da.:  and  at  7  per  cent,  in  10 
yrs.  2  mo.  27  da. 


How  do  we  find  the  amount  of  any  sum  by  the  table? 


DUODECIMALS.  299 

SECTION  XIV. 

DUODECIMALS* 

Obs.  1,  Duodecimals  are  fractions  of  a  foot ;  or  a  species  oj 
nuinbers,  of  which,  the  ratio  of  increase  and  decrease,  is  twelve.  The 
term  is  derived  from  the  Latin  numeral  duodecim,  which  signifies 
twelve. 

The  denominations  are  Feet,  Inches  or  Primes,  Seconds,  Thirds, 
Fourths,  &c. 

TAi3LE. 

12  fourths  ("")  make  1  third,         marked  '". 

12  thirds,  *'       1  second,            ** 

12  seconds,  **       1  inch  or  prime, ''in.or'. 

12  inches,  *•       1  foot,                 '*         ft. 

Remai.k  1.  The  marks  ',  ",  '",  "",  &c.,  which  distinguish  the 
different  denominations,  are  called  indices.  The  foot  has  no  index, 
being  considered  as  the  unit. 

2.  Duodecimals  may  be  added  Or  subtracted  in  the  same  manner 
as  other  compound  numbers.     (Sect.  IX,  Arts.  4  and  5.) 

MULTIPLICATION  OF  DUODECIMALS. 

Obs.  2.  Duodecimals  are  used  principally  in  measuring  surfaces 
ox  in  finding  the  solidity  of  bodies .  The  former  is  ascertained  by 
multiplying  together  the  length  and  breadth.  (Sect.  IX,  Art.  2, 
Obs.  15.)  And  the  latter  is  found  by  multiplying  together  the 
length,  breadth  and  thickness.     (Sect.  IX,  Art.  2  Obs.  18.) 

Remark  1.  As  12  inches  make  1  foot,  12  "  make  1  ',  &c.,  it  fol- 
lows that 

1  '  is         -  -  *  -  -         y  2  of  a  foot. 

1  "  is  y  2  of  an  inch,  or  ~  of  yV  =       -  ttt  of  9,  foot. 

1  '"  is  yV  of  1  ",or  yV  of  VVof  f  ',or  -^V  of  y\  of  yV=  TT2T  of  a  ft. 
1  ""  is  yV  of  1  '",  or  yV  of  V2  of  1  ",  or  y^  of  ,-V  of  tV  of  1 ',  or 

yV  of    yV  of    yV   Of    y^    =      o  oT3  6    Of    »   fOOt,   &C. 

The  foot  being  regarded  as  the  unit,  and  the  other  denominations 
as  the  fractional  parts  of  this  unit,  or  foot,    it  follows,   that  if  we 

What  are  duodecimals?  From  what  is  the  term  derived?  What  are  the  de- 
nominations? Repeat  the  table.  What  are  the  marks  ',"j"',  "",  &c.,  called 
Why  has  the  foot  no  index?  How  may  duodecimals  be  added  and  subtract- 
ed? For  what  are  duodecimals  principally  used'?  How  is  the  former  ascer- 
taiued?  How  is  the  latter  fouud?  What  part  of  a  foot  is  1  '?  What  part  of 
afoot  is  1  "?  Is  1  '^?  Is  ""'  If  w^  multiply  feet  by  feet,  what  shall  we 
obtain  as  the  reault?    Why? 


/^OO  COMMON    ARITHMETIC.  Scct.    XIV 

multiply  feet  by  feet,  we  shall  obtain  feet  as  the  result ;  as  1  X  1 
=  1.  But  if  we  multiply  feet  by  inches  {')  or  inches  (')  by  feet, 
the  result  will  be  inches  (')  because  1  inch  is  yV  of  a  foot,  ai:d  1  X 
J-  =  J-  =  1  ' 

Also,  inches  (')  multiplied  by  inches  (')  produce  seconds,  (")  as 

12     y\     12    14  4    ^  ^    ^  • 

Again,  feet  multiplied  by  seconds  (")  or  seconds  (")  by  feet,  pro- 
duce seconds,  (")  as  1  X  ttt  =  ttt  =  1  "• 

And  inches  (')  multiplied  by  seconds,  (")  or  seconds  (")  by  inches 
(')  produce  thirds,  ('")  as  ^-^  X  titt  =  ttVt  =  1  '"• 

And  seconds  (")  multiplied  by  seconds  (")  produce  fourths,  ("")  , 
as  TTT  X  T¥4  =  2  0T3  6  =  1  '"'.— Hencc— 

Obs.  3.     TheiJrodiict  of  any  two  denominations  is  always  of  that 
f  enomination,  denoted  by  ike  stem  of  their  indices. 

Ex.  1.     How  many  square  feet  in  a  room  14  ft.  8'   10"  in  length, 
and  12  ft.  7'  9"  in  width? 

Ans.   186  sq.  ft.  4' 2"  5'"  6'"'. 

1st  Operation. 
14  ft.  8'  10" 
12  ft.  7'     9" 


Ans. 


8  ft. 
176  ft. 

11'  0"      7'"  6"" 
7'   1"     10"' 
10'  0" 

186  ft. 

4'  2"      5"'  6"" 

2d   Operatioyi. 
14  ft.     8'   10" 

12  ft.     7'     9" 

176  ft.   10'     0" 
8  ft.     7'     1"   10"' 

11'     0"     7'"  6"" 

'or 


Ans.        186  ft.    4'     2"     5'"  6"" 

In  the  1st  operation,  we  multiply  first  by  9".   10"  X  9"  =  90"" 
(Obs.  3.)  90""  -i-  12  =  7'"  6"".     We  set  down  6""  and  carry  7"' 


Feet  and  inches  multiplied  together,  produce  what?  Why?  Inches  into 
nehes produce  what?  Why?  Feet  into  secon^^s  produce  what?  Why? — 
iiJCJies  into  seconds  produce  what?  Why?  Seconds  into  seconds  produce 
what?  Why?  Of  what  denomination  is  the  product  of  any  two  denomina- 
Hons?     In  the  operations  of  Ex.  1,  how  are  the  partial  products  written? 


DUODECIMALS.  301 

8'  X  9"  =  72'"  and  1'"  to  carry  m-ke  19'".  79'" -^12=  6"  7"V 
We  set  down  7'"  and  carry  6".  \\  ft.  X  9"  =  126",  and  6  to  car- 
ry make  132".  132"-^:2=£  11'  0",  which  we  set  down.  We 
multiply  by  7'  and  12  ft.,  setting  down  and  carrying  in  the  same 
manner.  Finally,  we  add  the  several  products  together,  observing 
to  carry  ]  for  every  12,  as  in  multiplying. 

In  the  2nd  operation,  we  multiply  first  by  12  ft.,  and  afterwards 
by  7',  and  9"  seLti  ig  down  and  carrying  as  in  the  first  operation. 

Remark. — Tn  both  th?pe  operations  the  learner  will  perceive  that  we  place 
those  numbers  of  the  same  denomination  under  each  other. 

From  the  preceding  remarks  we  derive  the  following 

Rule.     I.      Write  ike  muUipUer  under  the  multiplicand. 

II.  Multiply  each  denomination  of  the  multiplicand  {commencing 
at  the  right,)  by  each  denomination  of  the  multiplier  and  write  each  pro- 
duct under  i  s  corresponding  denomination. 

III.  Finally,  add  together  the  several  partial  products,  as  they 
stand,  carrying  for  every  19,  both  in  multiplying  and  adding. 

Remark  1.  The  learner  will  observe  that  feet  multiplied  by  feet  produce 
square  feet,  and  the  same  remark  is  observed  of  all  the  denomiuations  of  Zi»far 
measure.  (Sect.  IX,  Art.  *2,  Obs.  15.  Rem).  Also,  square  measure  multiplied 
by  linear  measure  produces  solid,  or  cublic  measure. 

2.  The  inches  in  duodecimals  are  usually  called  Carpenters^  inches.  They 
are  i\e\iher  linear,  square,  nor  cubic.  In  measuring  Zen^<A,  an  incA  is  one-twelfth 
of  a  foot.  In  measuring  surfaces,  an  inch  is  one-twelfih  of  a  square  foot;  and 
measures  1  foot  in  length,  and  1  inch  in  wilth.  In  measuring  solids,  an  inch  is 
one-twelfih  of  a  solid,  or  cubic  foot;  and  measures  I  foot  in  length,  1  foot  in 
thickness,  and  1  inch  in  width. 

2.  How  many  square  feet  in  12  boards,  each  18  ft.  7'  6"  long, 
and  1  ft.  4'  wide.  Ans.  298. 

3.  How  many  square  feet  in  a  room  24  ft.6'  9"  in  length,  and  18 
ft.  9'  10"  in  width?  Ans.  462  sq.  ft.  3'  0"  4'"  6"", 

4.  How  many  solid  feet  in  a  vat  9  ft.  6'  long,  7  ft.  8'  wide,  and  6 
ft.  9'  deep?  Ans.  491  sq.  ft.  7'  6". 

5.  How  many  solid  feet  in  a  box,  17  ft.  8'  4"  in  length,  6  ft.  7' 
3"  in  width,  and  4  ft.  13'  6"  in  depthl 

Ans.  530  sq.  ft.  8'  8"  6"'  2""  6"'". 

Obs.  4.  Painters',  Pavers'  or  Plasterers'  work  is  generally  com- 
puted by  the  square  yard.  Square  feet  are  reduced  to  square  yards 
by  dividing  by  9.     (See  table,  square  measure.) 

What  is  the  rule?  What  does  a  linear  measure  multiplied  by  linear  measure 
produce?  What  does  square  measure  multiplied  by  linear  measure  produce? 
What  are  the  inches  in  iiuodecimals  usually  cnlled?  What  is  an  inch,  and 
how  much  does  it  measure  in  measuring  lengths,  surfaces  and  solids?  How  is 
painters'  pavers  and  plasterers'  work  computed? 


yC2  COMxMdN    AAlTHiMETlC.  Scct.    XIV 

G.  How  mnny  square  yards  in  the  walls  of  a  room  which  measures 
'61  It.  4'  in  compass,  and  11  ft.  3'  in  height?     And  what  will  it  cost 
to  paint  the  walls  at  12-J  cents  per  square  yard? 

Ans.  76|  sq.  yds.;  cost  ^9.58~. 
?.  llow  much  will  it  cost  to  paint  a  building  measuring  96  ft.    9' 
in  compass,  and  15  ft.  8'  in  height,  at  10  cents  per  square' vard? 

Ans.  816,84-i-. 

8.  How  much  will  it  cost  to  pave  a  side  walk  124  ft,  G'  in  length, 
and  7  ft.  10'  in  width,  at  25  cents  per  square  yard? 

Ans.  ^27.09. 

9.  How  much  will  it  cost  to  pave  a  yard  35  ft.  G'  in  length,  and 
26  ft.   8'  in  widtli,  at  80  cents  per  square  yard.^ 

Ans.  830. 37|. 

10.  How  miicli  will  it  cost  to  plaster  a  ceiling,  at  18  cents  per 
square  yard,  it  measuring  '21  ft.  8'  in  len'^th,  and  12  ft.  9'  in  width? 

Ans.  ^5.52^. 

11.  How  mucti  will  it  cost  to  piaster  a  ceiling  18  ft.  G'  in  length, 
and  10  ft.  2'  in  width,  at  20  cents  per  square  yard? 

Ans.  $4.18  nearly. 

12.  A  certain  room  measures  25  ft*  3'  in  length,  18  ft.  4' in 
width,  and  9  ft.  in  heighth.  Required  the  cost  of  plastering  it,  at 
26  cents  per  square  yard,  deducting  3  doors,  eac!i  6  ft.  G'  by  2  ft.  8', 
a  fire  place  6  ft.  6'  by  4  ft.  4',  and  4  windows,  each  5  ft.  8'  by  3  ft. 
3'?  '  Ans.  $32,032. 

13.  A  certain  building  is  three  stories  in  height;  each  story  has 
12  windows,  each  6  ft.  10'  by  3  ft.  6'.  Required  the  cost  of  the 
glazing  at  1 6  cents  per  square  foot.  Ans.  $129.15. 

Obs.  5.  Some  parts  of  Carpenters'  and  Joiners'  work  are  compu- 
ted by  the  sq.  yd.;  other  parts  such  as  flooring,  &c.,  are  computed 
by  the  square.     A  square  consists  of  100  square  feet. 

14.  How  much  will  it  cost  to  lay  a  floor  22  ft.  6'  in  length,  and 
18  ft.  6'  in  width,  at  $3.50  per  square?  Ans.  $14.56j. 

15.  How  much  will  it  cost  to  lay  a  floor  30  ft.  8'  in  length,  and 
26  ft.  6'  in  width,  at  $4.50 per  square,  deducting  a  place  for  the 
stairs  9  ft.  10'  by  4  ft.  G';  and  a  fire  place  5  ft.  10'  by  4  ft.  6'. 

,  Ans.  833.39|. 

Obs.  6.  Masons'  work  is  sometimes  computed  by  the  solid  foot 
and  at  other  times  by  the  i:)erc}i.     A  perch   measures    16|-   feet   in 


How  are  square  feet  reduced  to  square  yards?  How  is  flooring,  roefingf, 
&c.,  computed?  Of  how  much  does  a  squar'e  (ionsist?  How  is  masons'  work 
sometimes  con\p«ted? 


fOWEftS     ANtJ    ROOT.  £03 

letigth,  1^  ft.  in  breadth,  and  1  foot  in  dcptli,  and  contains  IC^  X 
1^X1=  24|  solid  feet.     (Sect.  IX,  Art.  2,   Obs.  18.)     Hence— 

To  find  the  number  of  perches  in  any  wall  or  solid  body : 

Obs.  7.     Find  the  contents  in  solid  feet,  and  divide  this  by  24  J. 

16.  How  man}^  perches  in  a  wall  72  ft.  3'  long,  12  ft,  8'  high,  and 
2  ft.  thick?  Ans.  73.952+. 

17.  How  many  perches  in  a  wall  48  ft.  10'  long,  15  ft.  6'  high, 
.and  2  ft.  3'  thick?     What  will  it  cost  to  build  it  at  Ji^'2.50per  perch f 

Ans.  68.81  +  perches.      Cost  $172.U25. 

18.  How  many  bricks  in  a  wall  100  ft.  4' long,  10  ft.  6'  high, 
and  1  f .  8'  thick,  allowing  20  bricks  to  the  solid  foot,  and  how 
much  will  it  cost  to  lay  this  wall,  at  '$Q  per  thousand  bricks? 

Ans.  35116  bricks .     Cost  §?2  i  0.70. 

19.  How  many  bricks,  each  8  inches  long,  4  inches  wide,  and  2 
inches  thick  will  it  take  to  build  a  wall  40ft.^6'  long,  14  ft.  4'  high, 
and  2  ft.  8'  thick?  Ans.  417961 

20.  A  certain  brick  building  is  28  ft.  6'  in  leng-th,  24  ft.  6'  in  width, 
and  15  ft.  9'inheighth,  and  the  walls  are  1  foot  thick.  Now  de- 
ducting 4  doors,  each  6  ft.  9' by  2  ft,  8'  and  10  windows,  each  6  ft. 
6'  by  3  ft.  6',  how  many  bricks,  each  8'  long,  4'  wide,  and  2'  thick, 
did  it  require  for  this  building,  and  how  much  did  they  cost  at  ^^3.25 
per  thousand? 

Ans.  It  took  36990  bricks,  and  they  cost  ^120.2 If 


SECTION  xy. 


POWERS  AND  ROOTS. 


Def,  1 .  A  Power  of  any  number,  la  theiiroduct  arising  frotii  mii- 
tiplying  it  into  itself.  Thus  ;  4  is  a  power  of  2,  because  2  X  2  ::=4; 
and  27  is  a  power  of  3,  because  3  X  3  X  3  =  27,  <fec. 

Def.  2.  A  Root  of  any  number,  is  a  number  which  multiplied  into 
itself  a  certain  number  of  times  will  produce  the  given  number.  Thus: 
2  is  a  root  of  4,  because  2X2=4;  3  is  a  root  of  27,  because  3X 
3X3  —  27;  and  4  is  a  root  of  256,  because  4X4X4X4  = 
256. 


What  does  a  perch  measure?  How  many  solid  feet  does  it  contain?  How 
do  we  find  the  Dumber  of  perche^s  iu  a  wail  or  solid  body?  What  is  a  power? 
Give  examples.     What  is  a  roDt?     Give  examples? 


304  COMMON    ARItHMETIC.  Scct.    XV 

ARTICLE  1.     Involution. 

Obs.  1.  Involution  is  the  method  of  Jindlng  the  powers  of  num- 
bers. 

Obs.  2.  Powers  are  divided  into  several  orders,  called  the  first, 
second,  third,  fourth  power,  &c.  .  " 

Obs.  3.  The  number  wliicli  is  to  be  involved  is  called  the  root, 
OY  first  power.  The  other  powers  derive  their  name  iromthe  number 
of  times  the  root  is  emp)loy€d  as  a  factor,  in  producing  this  power, 
thus  : 


2  is  the  first  power  of 

2X2=  4  is  the  second  power 

2X2X2=   8  is  the  third  power  o: 


of  2. 

Tkit  of   2. 

-^x\-^/\- -^'^  ^"^  vw.i^.  power  of  2. 

2X2X2X2^16  is  the  foarth  power  of  2. 


Remark. — 'The  second  power  of  a  number  is  usually  called  the  42'"^^^ 
The  third  power  is  usually  called  tiie  cube.: 
The  fourth  power  is  usually  called  the  biquadrate: 
The  fifth  power  is  usually  called  the  sursolid: 

The  other  powers  generally  receive  no  other  name  tlian  their  numeral  dis- 
tinction, as  the  sixthpower,  the  seventhpower,  &c.  of  the  root,  or  first  power. 

MENTAL    EXERCISES, 

1.  Whatis  the  square  of  3?    Of  4;  5;  6;  8;   11;  7;   lOfO;   12? 

2.  What   s  the  cube  of  3?     Of  4;  7;  6;  5;   12.  9;   11;   10;  8? 

3.  What  is  the  fourth  power,  or  biquadrate,  of  3?     Of  5;  4;  6? 

4.  What  is  the  fifth   power,  or  sursolid,  of  2?     Of  3;  4;  5*  7? 
6.  What  is  the  sixth  power  of  2?     Of  3? 

6.  What  is  the  seventh  power  of  2?     Of  3? 

Obs.  4.  The  number  that  denotes  the  power  to  which  the  root  is 
to  be  raised  is  called  the  index,  or  exjjonent  of  the  power,  and  is  usu- 
ally written  at  the  right,  and  a  little  above  the  root.     Thus  : 

3*  signifies  3,  or  the frst  power  of  3. 

3^  signifies  3X3=9,  or  the  square  of  3. 

3^  signifies  3X3X3=27,  or  the  cube  of  3. 

3"*  signifies  3X3X3X3=81,  or  the  biquadrate  of  3,  &c.,  &c. 

Remark  1. — To  find  the  second  power,  or  square  of  3,  we  use  3  as  a  facto  T^ 
twice  ;  ihat  is,  we  multiply  it  into  itself  once. 

Thus:  3X3=9. 

What  is  Involution?  How  are  powers  divided?  What  is  the  number  to  be 
involved  called?  From  what  do  the  other  powers  receive  their  name?  Give 
examples.  What  is  the  second  power  of  a  number  usually  called?  The  third 
power?  The  fourth  power?  The  fifth  power?  What  name  do  the  other  pow- 
ers receive?  What  is  the  number  denoting  the  power  to  which  the  root  is  to 
be  involved  called? 


Art.   1.  iNvoLUT[ON,  305 

2. — To  find  the  cube,  or  third  power  of  3,  we  use  3  as  a  factor  three  times; 
that  is,  we  multiply  it  into  itself  twice. 

Thus  :  3  X  3  X  3  =  27. 

3. — To  find  the  biquadrate,  or  fourth  power  of  3,  we  use  3  as  a  factor  four 
times;  that  is,  we  multiply  it  into  itself  three  iiines. 

Thus:  3X  3  X  3  X  3  =  81.     Hence— 

To  involve  any  number  to  any  required  power  : 

Obs  5.  Multiply  (he  given  number  into  itself,  until  it  has  been  em- 
ployed as  a  factor  as  many  times  as  there  are  units  in  the  index,  de- 
noting the  power  to  lohich  it  is  to  be  raised. 

Remauk. — The  number  of  multiplications  in  involving  any  number  to  a  re- 
quired power,  is  always  1  less  thnn  the  index.  Thus,  3^  signifirs  that  3  is  to  be 
employed  as  a.  factor  three  times,  but  to  be  multiplied  into  itself  but  twice. 

Thus  :  3  X  3  X  3  =  27,  &c. 

Obs.  6,  All  the  powers  of  1  are  the  same — that  is,  1.  For  1^ 
=  1  ;   1^=3  1,  &c. 

EXERCISES    FOR   THE    SLATE. 

1.  What  is  the  square  of  14?  Ans.   196. 

2.  What  is  the  cube  of  13?  Ans.  2197. 

3.  What  is  the  fourth  power  of  1 5?  Ans.  50625. 

4.  How  much  is  13^?     14^?     18^?  20^?     30^?     40«? 

,         .         ,  S      169;     2744;     324;     160000;     2430C000; 

Ans.  m  order.       ^  4096000000. 

9  is  the  square  of  3.  To  multiply  9  by  3,  and  that  product  by 
3,  we  obtain  81  =  3%  which  is  the  same  as  9  X  9,  or  3"  X  3^. 

3'  =  27  ;  27  X  3  X  3  =  243  =  3%  which  is  the  same  as  27  X 
9,  or  3^  X  3^ 

3^  =  27  ;  27  X  3  X  3  X  3  =  729  ==  3%  which  is  the  same  as 
27  X  27,  or  3^  X  3^  &c. 

Obs.  7.  Hence — The  fourth  p)ower  is  equal  to  the  square  mul- 
tiplied by  the  square. 

The  fifth  poiver  is  equal  to  the  cube  multiplied  by  the  square. 

The  sixth  power  is  equal  to  the  cube  midtipUed  by  the  cube. 

The  seventh  power  is  equal  to  the  fourth  poioer  multiplied  by  the 
cube,  d'c. 

That  is — The  product  of  any  two  powers  of  a  nwnber,  is  equal 

How  do  we  involve  nu  nbers  to  any  required  power?  How  many  times  is  a 
number  used  as  a  multiplinr  in  involving  it  to  any  required  power?  VVliut  are 
all  the  powers  of  l?  To  what  is  the  fourth  power  ofaiiy  number  equal?  Tho 
filth  power?  Tlie  sixth  power?  The  seventh  power?  To  what  is  the  pro- 
duct of  any  two  powers  of  a  number  equal  ? 


306  COMMON    ARITHMETIC.  Sect.    XV 

to  that  ■power  of  the  numher  denoted  ly  the  swn  of  their  exponents. 

From  the  above  illustrations,  the  following  proposition  is  also  evi- 
dent : 

Obs.  8.  The  quotient  arising  from  dividing  any  i-xiiuer  of  a 
numher  hy  another  jjower  of  the  same  numher,  is  equal  to  that  jjower 
of  the  number  denoted  by  the  difference  of  their  exponents. 

Let  the  learner  find  the  results  of  the  following  questions  by  the 
wo  preceding  observations  : 

5.  Required— the  fourth  power  of  16?      18?     22?     2.5? 

Ans.  in  o  <:^.qy.     65536;   104976;  234256;  3906525. 

6.  Required—the  fifth  power  of  17?     19?     21?     24?     30? 

A  ,      '         ,       S      1419857;  2476099.  4084101;  7962624; 

^^''''  "^  "'■"^■^'-  \        243o:]boo. 

7.  Required— the  sixth  power  of  16?     18?     22?     25? 

Ans    in  order"    \      16777216  ;     34012224 ;     113379904 
Ans.  m  OHici.     ^  244140625. 

8.  Required— the   seventh   power  of  17?     19?     21?     24?     30? 

An^    in  nH        ^410338673;     893871739;      1801088541; 
\      4586471424;  21870000000. 
7.  Divide  the  ninth  power  of  12  by  the  seventh  power  of  12. 

Ans.   144. 

10.  Divide  14-'  by  14^  Ans.   196  =  14". 

11.  Divide  18'^ 'by  18'.  Ans.   5832=  18^ 

12.  Divide  36'^'^  by36^^  Ans.  60466176. 

Obs.  9.  Fractions  may  be  involved  as  w^eli  as  whole  numbers, 
by  involving  loth  the  numerator  and  denominator  to  the  power  indi- 
cat'xl  hy  the  exponent. 

Remark  1. — A  fraction  to  be  involved  is  generally  written  within 
a  parenthesis  ;  thus  :  (|)^.  This  signifies  that  f-  is  to  be  multiplied 
into  itself,  or  squared.  Thus  :  f-  X  |  =  -J  ;  and  {\Y  =^  \  X  |  Xl 
=  |i;  but|^=.|  =  2|:   and|3  ^^,k^. 

2. — Any  power  of  a  proper  fraction  is  less  than  the  fraction  itself 
This  is  evident  from  the  fact  that  the  numerator  is  less  than  the  de- 
nominator; (Sect.  Vlil.  Art.  1.  Obs.  11.)  and  therefore,  when  the 
fraction  \^  involved,  the  multiplier  of  the  denominator  is  ^r^«^er  than 
the  multiplier  of  the  numerator,  and  this  being  the  case,  the  ratio 

To  what  is  ihe  quotient  arit^lnfi'  from  dividincr  asiy  power  of  a  number,  by 
another  power  of  the  same  niUMber,  equal?  How  may  fractions  be  involved? 
Plow  is  a  fraction  to  be  ir.volvod  generally  written?  Wiiat  is  the  value  of  the 
power  of  a  proper  fraction,  cojupared  with  the  value  of  the-  fraction  itself? 
^^how   why  this  is  the  chss? 


Art.  i.  INVOLUTION  897 

between  the  terms  of  the  fraction  is  diminished  ;  hence,  as  the  ratio 
is  the  quotient  of  one  number  divided  by  another,  (Sect.  XII.  Art. 
1.  Obs.  1.)  and  the  value  of  the  fraction  answers  to  the  quotierd  in 
division,  (Sect.  VIII.  Art.  1.  Obs.  9.)  it  is  evident  that  if  the  quo- 
tisnt,  or  ratio,  is  diminished,  the  value  of  the  fraction  is  also  dimin- 
ished. 

3. — It  is  generally  most  convenlen'  to  reduce  mixed  numbers  t; 
improper  fractions,  or  the  fractional  part  to  a  de-AmaU  before  iavo'- 
ving  it  to  the  power  indicated  by  the  iniex.  Likewise,  co7/t;;o."/i.:/ 
and  complex  f ratio  AS  must  be  reiuced  to  simple  ones  befuiv  in- 
volving them. 

13,  What  is  the  square  of  J?  J?  J?  \\]  -^Kl  \-\]  l\'l 
^  of  I? 


lof  ^'^ 


14.   What  is   the   cube  off?     -^^l     -^jl     l-l-      tt-     Vi^     4: 
I3|j     '      '    '       '    '.       4    ■        {\\Y 


f2i~|  '  (74)"  2 

15.  How  mucins    |—  |     ?     j"?      f""'?       —  ?        —     l       (i 


of  1  of  A)-"? 

16.  What  is  the  square  of  25?  xVns.  Q'lb. 

Common    Operation.  ^ina lytic    Operati-jn. 

25  25=:20-f5 

25  204-  5 


1 25  1 00-[-25  =  Pro  t v."\  of  2  )-f  5  !> v     ^. 

50  400-f  100  =  Pro  i act.  ul  2J-J-5  by  20. 

Ans.   G25  Ans.   400-f-200+25  =  625=.h     Siiuare. 

Thus,  we  perceive  the  result  to  be  equal  to  the  square  of  the  first 
part,  (20,)^  plus  twice  the  product  of  the  two  part^  (20  X  5  X  2.) 
plus  the  square  of  the  second  part,  (5)".  The  operation  may  be 
contracted,  thus : 

25  5X5,  or  5^=25;  set  down  5  and  cany  2.     2X5=10; 

25  10X2=20,  and  2  to  carry  =  22;  set  down  2,    and  carry 

2.     2X2,  or  2^=4,  and  2to  carrv  =  6.  (^cct.  VI.  Art. 

625  2.  Ex.  13,     Hence— 

To  find  the  square  of  any  number  consisting  of  buL  tw^o  figures  : 
Obs.  10,      Square  the  right  hand  digit  for  the  unit  figure  of  the 

How  do  WH  proceed  with  mixed  numbers,  when  involviufj  thi'm?  With 
compound  and  conipl-'x  fractions?  liithe  analytic  operatioa  of  EIx.  16,  to 
what  is  liie  rosalt  fqiial?  How  theu  may  we  find  the  square  of  a  numb.^r  con- 
Kisiing  o£  but  two  figures? 


308  COMMON    ARITHMETIC.  ScCt.    XV 

result,  double  the  product  of  the  two  digits  for  the  ten's  figure  of 
the  result^  and  square  the  left  hand  digit  for  the  remaining  figures 
of  the  result,  observing  to  carry  in  each  case  as  in  Midtiplication 
of  Simple  Numbers. 

Note. — By  this  method  the  learner  can  often  find  the  square  of  numbers 
mentally,  which  would  otherwise  require  the  use  of  the  slate. 

17.  How  much  is  99^?     27^?     46^?     75^? 

Ans.  9801;  729;  2116;  5625. 

18.  How  much  is  92^?     86^?     50^?     lOOM 

19.  How  much  is  31=^?     44^?     .37''?     dO'?     60^?     10']    20"^? 

20.  How  much  is  120^?     144^?     300*^?     1728"?     256^? 

21.  How  much  is  14^?     9^"?     15^?    (4)^?    |o?     ill     (41)^? 


2J 

"22."  How  much  is  (1)'^?  (^)'?  .5^?  .07"?  .001' "?  4.75^  ? 
l.r? 

23.  How  much  is  .4^?      .02'?     1.0002^?     3.0702^?  (lOl)^? 

24.  How  much  is  7.9^?     1.16"?     211^?     200.002'?   .OOOl^? 

25.  How  much  is  (10|)^?     17. 9"?     216.01^?     100.01^? 

Article  2.     Evolution. 

Obs.  1.  Evolution  is  the  method  of  finding  the  roots  of  num- 
bers. 

Rem  AUK  1. — Evolution  is  the  opposite  of  Involution,  and  each  reciprocally 
proves  the  other.  By  the  one,  we  multiply  a  number  into  itself  to  find  the 
power:   by  the  other,  we  resolve  the  power  into  equal  faciors  to  find  the  root. 

2. — Different  roots  have  different  names,  corre.sponding  to  those  of  the  dif- 
ferent powers. 

Thus :  2  is  the  second  root  of  4,  because  2X2  =  4;  and  3  is 
the^hird  root  of  27,  because  3  X  3  X  3  =  27.     Hence — 

Obs.  2.  A  root  takes  its  name  from  the  number  of  times  it  is 
employed  as  a  factor  to  produce  the  given  power. 

Rb:mark  1. — The  second  root  is  usually  called  the  square  root. 
The  third  root  is  usually  culled  tlie  cube  root. 

The  otfier  roots  generally  receive  no  other  name  than  their  numeral  distinc- 
tion— as  the  /owr/A  roo^  Iht'  fifth  root,  &lc. 

Wiiat-^is  (1)6  advantage  of  this  rule?  What  is  Evolution  ?  VVliat  relation 
has  Evolution  io  Involution'?  How  is  lhi;5?  From  wh^it  do<?^  a  root  take  its 
name?  What  is  the  second  root  usually  called?  The  third  root?  What  name 
do  the  other  roots  generally  receive? 


Art.  2.  EVOLUTION.  309 

2. — Powers  and  roots  are  correlative  terms;  because,  if  one  number  is  a 
power  of  another,  the  latter  is  a  root  of  the  former.  Thus,  8  is  the  cubs 
of  2,  and  2  is  the  cube  of  8. 

MENTAL   EXERCISES. 

1.  What  is  the  square  root  of  9?     Of  16;  25;  49;  3G;  81;   100? 

2.  What  is  the  cube  root  of  8?     Of  27;  64;   125? 

3.  What  is  the  fourth  root  of  16?  Of  81? 

4.  What  is  the  square  root  of  144?     The  cube  root  of  729? 

Obs.  3.     This  character  (J)  is  the  sign  of  the  square  rcol. 

Other  roots  are  indicated  by  the  same  character,  with  the  index 
of  the  root  placed  above  it.  Thus  :  y  indicates  the  cube  root ;  J 
indicates  the  fourth  root,  &c. 

Remark  I.—  Roots  are  also  inlicated  by  a  common  fraction,  the  numerator 
of  which  is  1,  and  the  denominator  the  index  of  the  root.  Tliis  is  written  in 
the  same  manner  as  the  exponents  of  powers. 

J  \  \ 

Thus:  42  is  the  same  as  ^4  ;  8^  is  the  same  ^8  ;  16*  is  the 
same  as  ^16,  &c. 

2. — As  the  square  roultiplied  by  the  square  produces  the  fourth  power, 
(Art.  1.  Obs.  7.)  it  is  evident  that  the  square  root  of  the  square  root  is  the  fourth 
root. 

Thus:   l/\6=  4  ;    ^4  =  2  =  ^16. 

Likewise,  the  sixth  root  is  equal  to  the  square  root  of  the  cube  root,  or  the 
cube  root  of  the  square  root. 

Thus:  ^64  =  4;  ^4=2=^64.  And  ^64  =  8  ;  ^8  =  2 
=  764.     Hence— 

Obs.  4.  When  the  index  of  the  root  can  he  resolved  into  fac- 
tors, these  factors  denote  the  roots,  which  being  successively  found , 
will  give  the  required  root. 

Obs-  5,  The  process  of  finding  the  roots  of  numbers  is  called 
the  Extraction  of  Roots.  It  consists  in  finding  such  a  number,  as 
multiplied  into  itself  a  certain  number  of  times  w  11  produce  the 
power.     (Definition  2.) 

5.  How  much  is  ^144?     ^81;  i/216;  y2>U;  ^81? 

What  kind  of  terms  are  powers  and  roots?  Why?  What  is  the  sijrn  of 
the  square  root?  How  are  other  roots  indicated?  By  what  other  method  are 
roots  indicated?  How  is  this  fr>iction  written?  To  what  is  the  square  multi- 
plied by  the  square  equal?  To  what  is  the  square  root  pf  the  square  root  equal? 
To  what  is  the  sixth  root  equal?  When  the  index  of  root  can  be  resolved  into 
factors,  what  do  these  factors  denote?  What  is  the  process  of  finding  the  roots 
of  num'beirs  called?    In  what  does  it  consist? 


JfiO  coMviuN  A.-.ini.MEric.  Sect  XV 

G.   How  much  b  ^j  21?     ^100:  '1/5X2;  ^^29;   i^243?    '"" 

7.  How  much  is   764?     ^30;  l/\25;   ^16;   ij64;   yi2d1 
Obs.  7.      /tooA9  o/'  FrudloriJi  may  be   extracted,  by  iinding  the 

roofs  of  both  the  numerator  and  the  denominator. 

V4                        5 
Tluis  :   ^4  =  I;   ^-.y".  =  | ;  but =f ;  and =|=2t,,  &c. 

7        _  i/S       .       " 

(1.  Mixed  rmvihcrs  must  be  reduced  to  impr  over  /'^'actions  before 
extracting  tlieir  roots. 

8.  How  much  is  ^/ At, ?     V^i^;  ^/^;  ^/Vf^? 

1 

9.  How  much  is  ^/\V\     714^   <i^V;     • 

^5/64 

10.  How  much  is  ^12^-?      v^l3i;   ^3|;   ^^2^f? 

Obi.  8.     Yvi>7i\  d]es(=  rxamples  we  notice  tliis  consideration  : 

Tf  the  power  con h' his  a  proper  fraction,  the  root  also  contains  a 

proper  fraction.     And  con\'er:.^e!y  : 

a.     If  the  root  contains  a  proper  fraction,  any  power  of  this  root 

m.itsi  also  contain  a  proper  fraction.     This  is  evident  from  Art,   1. 

Obs.  9.  Rem.  2.     Also  : 

Ob.s.  9.  Jf  the  power  contains  a  decimal,  its  root  must  also  co7i~ 
tain  a  decimal.     And  conversely — 

a.  Jf  the  root  contains  a  decimal,  any  power  of  tJds  root  must 
contain  a  decimal. 

This  is  evident  from  the  fact,  that  if  we  multiply  by  a  decimal, 
the  product  must  contain  a  de  imal.  (Sect.  VIII.  Art.  11.) 

Rkaiakk. — From  the  f.  ct  that  the  product  must  contain  as  many  deci- 
mals as  both  factors,  it  is  e/idcnt  ihal  the  root  must  contain  one  half  as  many 
decimals  as  its  square,  one  third  as  rnamj  decimals  as  its  cube,  ^c. 

11.  How  much  is  ^/. 25?     x/.64?     -^.027? 


12.  How  much  is  ^y.343?     ^.729?     ^.16X^'^     -^30—3? 

Obs.  lO.  A  number  whose  root  can  be  exactly  found,  is 
called  ^  perfect  power ^  and  its  root  is  called  a  rational  number.  Thus  : 
9,  \Q,  25,  27,  &c.,  are  perfect  powers,  and  their  roots,  3,  4,  5,  and 
3,  (the  cube  root  of  27,)  are  rational  numbers. 


How  may  roots  of  fractions  be  extracted?  How  do  we  proceed  with  mixed 
nnmbers  when  tliey  occur'?  v\  liat  consideration  is  noticed  from  Ex.  8,  9,  and 
10?  Show  why  thisis«orrect?  What  is  said  respectinc];  decimals  in  the  power 
or  root?  Show  why  tlii*!  is  correct.  H;w  many  decimals  must  the  root  con- 
tain, compared  with  those  of  the  square?  Of  its  cube?  Show  why  this  is 
correct.     What  is  a  perfect  power?     A  rational  number? 


Art.  2.  EVOLUTION*  311 

Obs.  11.  A  number  wliose  root  carmot  be  exactly  ascertained,  is 
called  an  imperfect  power,  and  its  root  is  called  a  surd,  or  irrational 
number. 

Thus  :  11,  17,  45,  &c.,  are  imperfect  powers,  and  their  roots  3.3 
+,  4. 1-f-,  6.  7-|-  are  surds. 

Remake. — A  number  may  often  be  Vi  perfect  power  of  one  order,  and  an  im- 
perfect power  of  another  order.  Thus,  25  is  a  perfect  power  of  the  second 
order,  and  an  imperfect  power  of  the  third  order.  Also,  61  is  a  perfect  power 
of  the  second,  third,  <xud  Jifth  orders,  and  an  imperfect  power  of  the  fourth 
order. 

Obs.  12-  Every  root  and  power  of  1  is  the  same — thai  is,  1. 
Thus,  1%  P,  1%  v/1,  ^1,  l/i,  \J\,  &c.,  are  all  equal. 

Note. — The  preceding  exercises  in  Evolution  arc  designed  to  be  performed 
mentally  ;  but  it  frequently  happens  that  the  numbers  are  too  large  to  ascer- 
tain their  roots  in  this  manner,  and  we  have  to  use  the  slate.  In  such  cases 
there  is  a  particular  process  by  wliich  the  roots  are  found.  The  methods 
of  extracting  the  square  we  will  now  explain.  As  the  other  roots  seldom 
occur  in  practical  business,  they  are  not  treated  of  in  thi^  v»-ork. 

Case  1 . — Extraction  of  the  Square  Root. 

Obs.  13.  To  extract  the  square  root  of  a  niimler,  is  to  find  a 
number,  which  being  multiplied  into  Itself,  wilt  pr<^duce  the  given 
nurnber.  (Art.  1.  Obs.  4.  Rem.) 

Remark. — The  terms  Square  and  Square  Root  are  de- 
rived from  Geometry.  Tbe[sqrtare  may  be  considered 
as  a  square  figure,  made  up  of  several  smaller  square.'^, 
and  the  square  root  as  the  number  of  smaller  squarrs 
on  one  side  of  this  figure.  Thus,  in  the  diagram,  the 
large  square  is  made  up  of  16  smaller  ones,  and  there 
are  4  of  these  on  a  side  ;  therefore,  16  is  the  square, 
and  4  is  the  square  root.    Hence — 


Obs.  14.     The  extraction  of  the  Square  Boot  may  be  defined,  a 
finding  one  side  of  a  square,  when  the  square  contents  are  given. 
Obs.  15.     Take  the  following  numbers  : 

1,  2,  3,     4,    5,     6,     8,     9,      10,       99,       100. 
Their  factors  are     1,4,9,  16,25,36,  64,   81,   100,  9801,   10000. 
The  numbers  in  the  first  line  may  also  be  regarded  as  the  square 
roots  of  those  in  the  second  line.  (Obs.  2.  Rem.  2.) 

vVhat  is  an  imperfect  power?  A  surd  or  irrational  number?  Can  a  number  be 
a  perfect  power  of  one  order,  and  an  imperfect  power  of  another  order?  Give 
examples.  What  is  every  root  and  power  of  1?  What  is  it  to  extract  the 
square  root  of  a  number?  From  what  are  the  terms  square  and  square  root 
derUred?  What  may  the  square  be  considered?  The  square  root?  How  then 
may  the  extraction  of  the  square  root  be  defined! 


312  COMMON    ARITHMETIC.  Scct.  XV 

From  this  we  notice  the  following  considerations : 

a.  1st.    The  square  of  no  digit  exceeds  hoo  figures, 

b.  2nd.  When  a  number  contains  but  two  figures,  its  square  root 
contains  but  one.  Also :  When  the  number  contains  three  or  four 
l^laces  of  figures,  its  square  root  contains  2. 

c.  3rd.  The  square  of  no  number  contains  more  than  double 
the  nutnher  of  figures  in  the  number  squared,  and  but  1  less. 

Obs.  \Q.  Hence — When  we  extract  the  square  root  of  a  num- 
ber, we  first  point  it  off  into  periods  of  two  figures  each,  comme7icing 
at  the  right  hand.     By  this  means  we  ascertain — 

a.  1  St.  Bow  many  figures  the  left  hand  period  contains,  or  how 
many  figures  must  be  taken  to  find  the  first  figure  of  the  root. 
This  is  the  most  important  object  gained. 

b.  2nd.    Of  how  many  figures  the  root  will  consist. 

Remark. — From  Obs.  15,  c,  it  is  evident  that  the  square  root  must  con- 
tain as  many  figures  as  there  are  periods  in  the  square.  If  the  s^'^are  con- 
tains an  odd  number  of  figures,  the  root  will  contain  1  more  than  half  as 
many;  if  the  square  contaius  an  even  number  of  figures,  the  root  will  con- 
tain just  half  as  many.  Thus,  the  square  root  of  144  is  12;  and  of  6400, 
the  squa-e  root  is  80.  These  numbers  when  pointed  off  into  periods  stand 
thus:     i-*-*  ;   640  o. 

Ex,  1.  A  man  has  a  square  field  containing  625  square  rods. 
How  many  rods  does  it  measure  on  each  side?  Ans.  25. 

Operation. 

625  (25  \st  step. — In  this  example  we  wish  to  find  one 

4  side  of  a  square,  having  given  its  contents ;  that 

is,   we  wish  to  extract  the  square  root  of   625. 

45)225  (Obs.  14.)     We  first  point  off  the  given  number 

225  into  periods,  and  then  ascertain  that  the  left  hand 

period  contains  one  figure,  and  that  the  root  will 

000  cowt?im  two  figures.  (Obs.  16.)     'I  hese  two  figures 

(of  the  root)  of  course  are  a  ten  and  a  unit. 

2nd  step. — We  now  seek  the  greatest  square  in  the  left  hand  pe- 
riod, (6)  and  find  it  to  be  4,  the  square  root  of  which  is  2.  This 
we  place  at  the  right,  wi(h  a  Hie  to  separate  it  from  the  power. 

As  this  is  the  root  of  the  left  hand  period,  (6  hundreds)  it  is  in 
reality  2  tens,  (because  the  square  of  2  tens,  or  20,  is  4  hundreds, 
or  400,)  and  is  equal  to  one  side  of  a  square  measuring  2  tens 
(20)  rods  on  a  side.     This  may  be  shown  by  a  diagram  : 

"What  is  the  first  consideration  that  we  notice  from  examining  the  given 
numbers  and  their  squares?  Thesecond?  The  third?  What  is  the  first  thing 
we  do  in  extracting  the  square  root  of  a  number?  What  do  we  ascertain  by 
this  means?  How  many  figures  must  every  root  contain?  If  the  square  con- 
tains an  odd  number  of  figures,  how  many  figures  will  the  root  contain?  If 
the  square  centains  an  even  number  of  figures,  how  many  will  the  root  cijntain? 


Art.  2. 


EVOLUTION. 


^W 


Fig.  1. 


B 


20 
20 


400  sq.  rds. 


C 


The  contents  of  this  fig- 
ure we  find  to  be  20x20= 
400  sq.  rds.,  now  disposed 
of,  and  which  we  will  de- 
duct from  the  whole  num- 
ber of  rods  m  the  field, 
(.625)  showing  225  sq.  rds. 
remaining.  ^his  deduction 
is  readily  eff  oted  by  sub- 
tracting the  4  Hw^^^^As) 
the  square  of  the  2  (tens,) 
from  the  left  hand  period, 
6,  (hundreds,)  leaving  2 
(hundreds,)  and  bringing 
down  the  next  period,  (j45,) 


'^  D  20  rods. 

making  225  as  seen  in  the  operation. 

3d  step. — We  now  wish  to  enlarge  our  figue  by  the  addition  of 
225  sq,  rds.,  and  it  is  evident  that  this  addition  must  be  made  on 
two  sides,  in  order  to  preserve  the  square.  This  is  shown  in  the 
following  diagram  : 

FIGURE   2.  A/ 

20  rds. 5  rds. 


o 


20 
5 


B 


100  sq.  rds. 


20 
20 


400  sq.  rds. 


25  sq.  rds 


"D 


D 


20 
5 


100 sq.  yds 


20  rds. 


5  rds. 


Now  as  the  ad- 
dition is  made 
on  two  sides  of 
the  square,  if 
we  divide  the 
number  of  rods 
to  be  added, 
(225,)  by  the 
length  of  the 
two  sides,  (20 
+20=40,)  the 
quotient  will 
evidently  b  e 
the  width  of  the 
additions  made 
to  the  two  sides 
A  B,  B  C,  Fig. 
1 .  According- 
ly,  we  double 


In  the  operation  of  Ex.  1,  what  is  the  first  step?     What  do  we  ascertain  by 
this?     Whiit  is  the  second  step?     What  is  this  root?     Why?     To  what  is  it 
equal?    What  is  done  with  the  contents  of  this  figure? 
16 


314  COROION    ARITHMETIC.  Scct,    XV 

the  root  (2  tens,)  already  found,  making  4  (tens)  for  a  divisor, 
^th  step. — We  now  seek  how  many  times  our  divisor  (40)  is  con- 
tained in  our  dividend,  (225)  and  lind  it  to  be  contained  5  times  ; 
therefore,  5  rds.  is  the  width  of  the  additions,  and  40x^=200  ad- 
ditional rods  disposed  of,  and  25  rds.  remaining. 

5th  step. — But  looking  at  Fig.  2  we  perceive  that  there  is  still 
wanting  in  the  corner,  C,  a  small  square,  equal  in  size  to  the  width 
of  the  additions  at  the  sides,  (5  rds.)  As  this  is  a  square,  its  con- 
tents are  5X5=25  sq.  rds.,  which  completes  the  square,  and  gives 
25  rds.  as  the  length  of  one  side. 

If  the  learner  examines  Fig.  2  attentively,  he  will  perceive  the 
additions  measure  45  rds.  in  length,  (20  rds.  on  each  side,  and  5 
rds.  at  the  corner,)  and  5  rds.  in  width  ;  and  therefore,  if  we  multi- 
ply 45  by  5,  we  shall  obtain  the  same  result  (225.)  This  is  effected, 
in  our  operation,  by  rejecting  the  cipher  at  the  right  of  our  divisor, 
(40,)  and  also  the  right  hand  figure  (5)  of  our  dividend,  (225,)  and 
dividing  the  remaining  figure  of  our  dividend  (22)- by  the  remaining 
figure  (4)  of  the  divisor,  and  writmg  the  quotient  figure  both  in  the 
root,  and  also  at  the  right  of  our  divisor  (4.)  Next,  we  complete 
our  dividend,  (225.)  and  multiply  our  divisor  (45)  by  the  last  fig- 
ure of  the  root,  (5,)  and  subtract  the  product  from  our  dividend, 
which  in  this  case  leaves  no  remainder. 

Rejecting  a  figure  at  the  right  of  our  divisor  and  dividend,  is  in 
effect  dividing  both  by  10,  (Sect.  V.  Art.  4,  Obs.  1.)  and  this 
cannot  affect  the  result.  (Sect.  VI.  Art.  1.  Obs.  28.) 

Proof. — The  operation  may  be  proved  by  adding  together  the 
several  parts  of  the  last  diagram.     Thus  : 
The  square  A  contains  400  sq.  rds. 

The   additions  B  and  D  contain     200  sq.  rds.;  that  is,  100  sq.  rds. 
The  corner  C  contains  25  sq.  rds.  [each. 

And  the  entire  figure  contains    625  sq.  rds. 

Or,  by  Involution  :  25  X  25  =  625,  as  before.  (Art.  1.  Obs.  5.) 

Remark  2. — Perhaps  the  learner  has  inquired,  before  this  time — Why  do 
we  commence  at  the  right  to  point  off  the  given  nvmbrr  into  periods?     The  an- 


How  la  this  deduction  effected?  Wiiat  is  the  third  step?  On  how  many- 
sides  must  this  addition  be  made?  Why?  How  do  we  find  ihe  width  of  these 
additions?  How  do  we  obtain  our  divisor?  What  is  (he  fourth  step?  How 
are  the  contents  of  the  corner  C  ascertained?  What  is  tlie  length  of  all  the 
additions?  How  is  this  found?  What  is  their  width?  How  then  may  we 
find  the  contents  of  all  the  additions'  In  the  operation,  how  is  this  efTected? 
What  effect  does  it  iiave  upon  the  divisor  and  dividend  to  reject  a  figure  from 
the  right  of  each?  Does  this  effect  the  result?  Why  not?  How  do  we  prove 
the  operatiot?  By  what  other  method?  Why  do  we  commence  at  the  right 
to  point  off  the  given  sum  into  periods? 


Art.  2.  EVOLUTION.  315 

is — Because,  after  the  first  figure  of  the  root  is  obtained,  there  must  be  two 

figures  in  the  given  sum  for  every  additional  figure  in  the  root.  [Rem.  1] 
Hence,  we  must  point  otf  from  the  right,  as  th«  first  figure  of  the  root  is 
found  from  the  left  hand  period.  We  commence  our  operations  at  the  loft 
on  account  of  the  remainders  wliich  occur.  [Sect.  V.  Art.  2.  Rule  for 
Short  Division.  Rem.  1.] 

From  these  remarks  we  derive  the  following 

RULE  FOR  EXTRACTING  THE  SQUARE  ROOT. 

1.  Point  off  the  given  sum  into  periods  of  two  figures  each,  pla- 
cing a  dot  over  the  unit  figure,  another  over  the  hundreds,  and  so  on, 
(Obs.  16.) 

II.  Find  the  greatest  square  in  the  left  hand  period,  and  place 
its  root  at  the  right,  like  a  quotient  in  Division.  Subtract  the  square 
of  the  root  from  the  left  hand  period,  and  bring  down  the  first  figure 
of  the  next  period  for  a  partial  dividend. 

III,  Double  the  root  already  found  for  a  defective  divisor;  seek 
how  many  times  this  defective  divisor  is  contained  in  the  partial 
dividend,  and  place  the  result  in  the  root,  and  also  at  the  right  of 
the  defective  divisor  to  complete  it.  Likewise,  bring  down  the  re- 
maining figure  of  the  period  at  the  right  of  the  partial  dividend,  for 
a  true  dividend. 

lY.  Multiply  the  true  divisor  by  the  last  figtire  of  the  root,  sub- 
tract the  product  from  the  dividend,  and  to  the  right  of  the  remmn- 
der  bnng  down  the  first  figure  of  the  next  period  for  another  par- 
tial dividend. 

V.  Double  the  root  already  found  for  another  defective  divisor, 
with  which  proceed  as  before,  until  all  the  periods  have  been  brougM 
down  and  divided. 

VI.  If  at  any  time  the  defective  divisor  is  not  coniained  in  the 
partial  dividend,  write  a  cipher  in  the  root,  and  bring  down  the  re- 
maining figure  of  the  period,  together  with  the  first  figure  of  the 
next  period,  for  another  partial  dividend,  with  which  proceed  as 
before. 

Proof. — Multiply  the  root  into  itself,  and  if  the  product  is  equal 
to  the  given  sum,  the  work  is  correct. 

2.  What  is  the  square  root  of  41616?  Ans,  204. 

Operation.  Proof. 

41616(204  204 

4  204 

^  404)1616  816 

^  1616  408 


% 


0000  41616  (Art,  1.  Obs,  5.) 


3l6  COMMON    ARITHMETIC.  Scct.    XV 

The  operation  may  also  be  demonstrated  in  the  followino-  manner  : 

3.  What  is  the  square  root  of  1296?  Ans.  36. 

Operation.  Proof. 

1296(36  36  =  30+  6 

9  30+  6 


66)396  180-f36 

396  900+180 

000  900+360+36=1296.      >) 

(Art.  1.  Obs.  10.) 

15/  step. — ^We  find  the  greatest  square  in  the  left  hand  period  to 
be  9,  and  write  its  root  (3)  in  the  quotient,  subtract  the  square  (9) 
from  the  period  (12),  and  bring  down  the  next  period  for  a  dividend. 

2ni  step. — As  the  root  consists  of  but  two  figures,  the  second  fig- 
ure must  be  such,  that  twice  the  product  of  the  first  and  second 
terms,  together  with  the  square  of  the  second,  must  complete  the 
square  [Art.  1.  Obs.  10.];  therefore  we  divide  our  dividend  [396] 
by  twice  the  figure  of  the  root  already  foun^,  viz.:  3  [tens] X 2=: 
6  [tens],  or  60,  and  find  the  second  figure  of  the  root  to  be  6. 

3d  step. — We  then  add  the  last  figure  of  the  root  [6]  to  our  divi- 
sor [60],  making  QQ-\-Q=QQ,  (or  merely  write  the  6  in  the  place  of 
the  cipher,)  and  multiply  this  by  the  last  figure  of  the  root  [6],  and 
subtract  the  product  from  the  dividend. 

By  this  means  we  obtain  twice  the  product  of  the  two  terms,  and 
the  square  of  the  second  term  ;  because  60  is  twice  the  first  term, 
which  being  multiplied  by  6  gives  twice  the  product  of  the  two 
terras,  and  6,  multiplied  by  6,  gives  the  square  of  the  second  term. 
It  produces  the  same  result  to  multiply  QQ  by  6,  because  66=60+ 
6,  and  it  is  an  established  principle  in  mathematics  ihdiiif  equals  are 
multiplied  by  equals,  their  products  will  he  equal. 

Obs.  1 7.  From  examining  the  first  nine  digits  and  their  powers, 
we  perceive  \X\^i  every  perfect  square  must  end  toitkO,  1,  4,  5,  6,  or  9; 
because  the   right  hand  figure  of  any  root  must  he  one  of  the  nine 

Why  do  /ve  commence  at  the  left  to  extract  the  root?  What  is  the  first  step 
in  extracting  the  square  root?  The  second?  Third?  Fourth?  Fifth?  If  at 
any  time  the  defective  divisor  is  not  contained  in  the  partial  dividend,  how 
do  we  proceed?  How  do  we  prove  the  operation?  What  is  the  first  step  in 
the  operation  of  Ex.  3?  The  second?  The  third?  What  do  we  obtain  by 
this  method?  Explain  why  this  is  correct?  By  what  other  method  can  we^ 
obtain  the  same  result?  Why  is  this  correct?  What  consideration  do  we  no-^ 
tied  from  examining  the  first  nine  digits  and  their  squares?  Why  is  this  cor- 
rect? 


Art.  2.  EVOLUTION.^  81' 

digits^  or  a  cipher  ;  and  the  square  of  no  digit  ends  with  any  other 
figure  than  one  of  the  above. 

Again — as  no  digit,  multiplied  by  itself,  will  produce  0  as  the 
unit  figure  of  the  result,  if  the  right  hand  place  of  the  root  is  a  ci- 
pher, the  last  period  must  be  ciphers  ;  that  is,  if  a  perfect  power  must 
end  with  1,  4,  5,  6,  9,  or  00  ;  likewise,  if  a  perfect  power  ends  with 
00,  the  remaining  figures  must  be  a  perfect  power.  This  is  evident 
without  further  demonstration. 

Also — since  every  defective  divisor  is  an  even  number,  (it  being 
twice  the  root,)  the  product  of  this  defective  divisor,  multiplied  by 
5,  must  end  with  a  cipher  ;  therefore,  if  the  root  ends  with  5,  the  square 
must  end  with  25;  because  the  square  of  5  is  25,  and  the  2  occupies 
the  place  of  the  cipher  above  mentioned.  Hence,  every  perfect  square 
must  end  wHh  00,  1,  4,  25,  6,  or  9,  and  if  a  perfect  square  ends 
with  00,  the  remaining  figures  must  also  be  a  perfect  square. 

EXERCISES    FOR    I  HE    SLATE, 

1.  What  is  the  square  root  of  502681?  Ans.  709. 

2.  What  is  the  square  rout  of  15876?  Ans.   126. 

3.  What  is  the  square  root  of  75076?  Ans.  274. 

4.  What  is  the  square  root  of  18541636?  Ans.  4306. 

5.  What  is  the  square  root  of  81072016?  Ans.  9004. 

6.  What  is  the  square  root  of  73960000?  Ans.  8600. 

7.  What  is  the  square  root  of  174740720400?     Ans.  418020. 

Obs.  18.  If  there  is  a  remainder  after  the  last  period  has  been 
divided,  we  may  form  new  periods,  by  annexing  ciphers,  and  con- 
tinue the  root  to  decimals,  to  any  degree  of  exactness. 

The  learner  must  not  expect  to  find  an  exact  root  in  such  cases, 
however,  as  the  square  of  no  digit  ends  with  a  cipher  ;  but  we  can 
approximate  suflaciently  near  by  the  aid  of  decimals.  To  prove  the 
operation,  he  must  add  the  remainder  to  the  square  of  the  root. 

8.  Required,  the  square  root  of  8?  Ans,  2.828-|-. 

9.  Required,  the  square  root  of  590?  Ans.  24.2o9-{-, 
10.  Required,  the  square  root  of  46780?         Ans.  216.2864-. 

Obs.  19.     To  extract  the  square  root  of  decimals — 

Commence  at  the  hundredth's  place,  and  point  towards  the  right. 

The  square  of  no  digit  ends  with  a  cipher  ;  what  is  the  inference  deduced 
from  tliis  fact^  If  a  perfect  square  ends  with  00,  what  must  the  remaining^ 
figures  be?  Why?  What  is  every  defective  divisor?  Why?  Wliat  infer- 
ence is  deduced  from  this  fact?  If  the  root  ends  with  5,  witii  what  must  the 
square  end?  Why?  With  wliat  then  must  every  square  end?  If  there  is  a 
remainder  after  the  last  period  has  been  divided,  how  do  w«  proceed?  Can  we 
find  an  exact  root  in  such  cases?  Why  not?  How  can  we  Hpproximate  suffi- 
ciently r*ar?  How  prove  the  operation?  How  do  we  extract  the  square  root 
of  decimals? 


818  COMMON-   ARITHMETIC.  Scct.    XV 

When  the  power  consists  of  whole  numbers  and  of  decimals  : 

Point  both  ways  from  the  unit's  place. 

Remark. — The  reason  why  we  poiat  decimals  from  the  left  towards  the  right, 
is  evident  from  the  fact,  that  everij  perfect  square  must  contain  twice  as  many 
decimals  as  its  root.  (Obs.  9.  Rem.)  Therefore,  for  every  decimal  in  the  root, 
there  raust  be  two  decimals,  or  a  full  period,  in  the  square;  and  if  the  peri- 
ods are  not  complete,  a  cipher  must  be  annexed  to  the  decimal.  Hence — we 
must  commence  at  the  left  to  point  off  decimals;  because  if  we  prefix  a  ci- 
pher to  the  left  to  complete  the  period,  the  value  of  the  decimal  is  altered, 
whilst  if  a  cipher  is  annexed  to  the  right  it  is  not.  (Sect.  VIII.  Art.  7.  Obs. 
6  and  7.)  "'I 

11.  What  is  the  square  root  of  356.164?         Ans.   18.872+.  '' 

12.  What  is  the  squr.re  root  of  .0225?  Ans.   .15. 

13.  What  is  the  square  root  of  225.6004?  Ans.   15.02. 

14.  What  is  the  square  root  of  .0000163216?     Ans.    .00404. 

1 5.  What  is  the  square  root  of  18.211?         Ans.  4 .  2673 1  +. 

16.  What  is  the  square  root  of  1 .004030076? 

Ans.  1.002012-}-. 

b  s.  20.     To  extract  the  square  root  of  a  common  fraction  : 

Pirst  reduce  the  fraction  to  its  simplest  form ;  then  if  both  terms 
are  perfect  squares,  extract  the  square  root  of  each  (Obs.  7.);  but  if 
both  terms  are  not  perfect  squares,  reduce  it  to  a  decimal,  and  then 
extract  the  square  root,  according  to  Obs.  19. 

17.  What  is  the  square  root  of  |4|-?  Ans. 

18.  What  is  the  square  root  of  ^jl^^?  Ans,    ^l 

19.  What  is  the  square  root  of  ^-^-^'l  Ans.    .3535-|- 

20.  What  is  the  square  root  of  yllo^  -^^^s.  .16. 

21.  What  is  the  square  root  of  lOyg?     (Obs,   7.  a.) 

Ans.   V  =  3^. 

22.  What  is  the  square  root  of  18|?     (184  =  18  .  8.) 

Ans.  4.335-^. 

23.  What  is  the  square  root  of  200yV¥4?  ^i^s.   14/^. 

24.  If  an  army  of  1 86624  men  were  drawn  up  in  a  perfect  square, 
how  many  men  would  there  be  on  a  side?  Ans.  432. 

25.  A  certain  square  held  contains  86436  hills  of  corn.  How 
many  are  there  on  a  side?  Ans,  294. 

26.  A  company  of  men  paid  $2304  to  a  charitable  institution, 
each  man  paying  as  many  dollars  as  there  were  men  in  the  com- 
pany.    How  many  men  were  there  in  the  company?       Ans.  48. 

27.  A.  has  two  lots,  one  measuring-  65  rods  m  length,  and  40 
rods  in  width,  the  other  measurino-  70  rods  in  lenMh,  and  20  rods 

Wh(  n  the  power  consists  of  Loth  whola  numbers  and  decimals,  how  do  we 
proceed?     Why  do  we  point  decimals  from  the   left  hand   towards  the  right? 


24' 
1  l_ 

2' 


HoW  we  extract  the  square  root  of  a  common  fraction'-.  ^^ 


# 


Art.  2. 


EVOLUTION. 


319 


a  square 


in  width.     B.  offers  him  for  these,  a  square  lot  containing  as  many 
rods  as  both.     How  many  rods  square  is  B.'s  lot?  Ans,  60. 

28.  What  is  the  difference  between  a  square  half  foot,  and  half  a 
square  foot?  Ans.  36  sq.  in. 

29.  If  it  takes  4356  square  tiles  to  pave  a  square  floor,  how  many 
are  there  on  a  side?  Ans.  66, 

30.  A  nurseryman  wishes  to  set  out  1521  fruit  trees,  so  as  to 
form  a  perfect  square.     How  many  must  he  put  in  a  row? 

Ans.  39. 

Case  2. — ArpLicATiON  of  the  Square  Root.      Triangles. 

Obs,  21.  A  Triangle  is  a  figure  hounded  hy  three  straight 
lines. \ 

Obs.  22.     When  one  line  meets  another  so  as  to  form 
corner,  or  right  angle,  it  is  called  a  Right  Angled  Triangle. 

Thus  :  A  B  C  is  a  right  angled  trian- 
gle, of  which  the  right  angle  is  B.  The 
side  A  B  is  called  the  hase  ;  ihe  side  B  C 
is  called  ihe  perpendieular  ;  and  the  side 
A  C  is  called  the  hyj)othenuse.  The  hy- 
pothenuse  is  always  the  side  opposite  the 
rigid  angle;  the  hypothenuse  A  C  is  opposite  the  right  angle  B. 

Obs,  23,  In  every  right  angled  triangle,  the  square  describe  ^  on 
the  hyjiothenuse  is  equal  to  ihe  sum  of  the  squares  of  the  other  two 
sides. 

Thus  :  if  the  base  is  4  feet,  and  the  perpendicular  3  feet,  the  hy- 
pothenuse is  J4^  -\-  o'^  =5  feet. 


Base. 


This  point  is  also  illustrated 
by  the  adjoining  diagram. 


What  is  a  triangle?     A  right  angled  triangle?     Which  eide  is  thi  hypothe- 
nnse?^    To  what  is  the  sqaare  of  the  hypothenuse  equal? 

tThi»;.«ieiiiiiiion  has  reference  only  to  rectilinear  trian<rles;  a  curvilinea    Jnanff'e  is  bounded 
py  curved  lines.  ^ 


S2p  feOaBMQN    ARITHMETIC.  Scct.  XV 

From  this  we  derive  the  following  considerations  : 

a.  1  st.  If  we  add  the  square  of  the  base  to  the  square  of  the  per- 
pendicular, and  extract  the  square  root  of  their  sum,  the  result  will  be 
the  hypothenuse. 

b.  2nd.  If  we  subtract  the  square  of  the  base  from  the  square  of  the 
hypothenuse,  and  extract  the  square  root  of  the  remainder ^  the  result 
mil  be  the  perpendicular, 

c.  3d.  7/^  we  subtract  the  square  of  the  perpendicular  from  the 
square  of  the  hypothenuse,  and  extract  the  square  root  of  the  remainder, 
the  result  will  be  ^the  base. 

Remark  1. — When  the  base  and  perpendicular  are  equal,  the  hypothenuse 
may  be  found  by  multiplying  the  base  (or  perpendicular)  by  1.4142.  This  is 
the  hypothenuse  of  such  a  triangle,  the  equal  sides  of  which  are  unity,  or  1. 

2  — The  base  and  perpendicular  are  sometimes  called  the  legs  of  the  triangU 

EXERCISES    FOR    THE    SLATE. 

1.  A  room  measures  16  feet  in  length,  and  12  feet  in  width.  How 
long  is  a  line  that  will  just  reach  from  corner  to  comer? 

Operation. 
16^=266 
12^  =  144 

400(20  feet.  Ans. 
4 

00 

2.  A  ladder  30  feet  long  will  just  reach  the  top  of  a  wall  by 
placing  its  foot  24  ieet  distant.     Required — the  heighth  of  the  wall. 

Ans.   18  feet, 

3.  A  certain  pole  is  60  feet  high.  How  far  from  the  foot  of  it 
will  a  hne  100  feet  long  reach  the  ground,  by  being  fastened  at  the 
top?  Ans,  80  feet. 

4.  The  distance  between  the  foot  of  two  rafters  is  32  feet,  and 
the  heighth  of  the  ridge  above  the  plate  on  which  the  rafters  rest  is 
12  feet.     Required — the  length  of  the  rafters.  Ans.  20  feet. 

6.  If  a  square  field  contains  35  acres,  25  sq.  rds.,  how  many  rods 
(i  es  it  measure  on  a  side?  What  is  the  distance  from  the  centre  to 
fc  ch  corner?  Ans.  to  the  last.  53.033-}-  rds. 

6.  The  walls  of  a  certain  buildino-  are  32  feet  in  lenfjth,  30  feet 

How  then  do  we  find  the  hypothenuse  v\hen  the  other  two  sides  are  given? 
How  do  we  find  the  perpendicular  when  the  other  two  sides  are  given?  How 
do  we  find  the  base  when  the  other  sides  are  given? 


Art  2. 


EVOLUTION. 


321 


in  width,  and  24  feet  in  heighth.     Required — the  length  of  a  Ime 
that  will  connect  the  two  corners  farthest  distant  from  each  other. 

Ans.  50  feet. 

7.  Suppose  a  pole  60  feet  long  to  be  so  planted  between  two 
straight  trees,  as  to  reach  a  limb  of  one  48  feet  high,  and  without 
moving  the  foot,  to  reach  a  hmb  of  the  other  36  feet  high.  How 
far  apart  are  the  trees?  Ans.  84  feet. 

Obs.  24.  A  mean  proportional  between  two  numbers  is  found  by 
extracting  the  square  root  of  their  product. 

Thus,  a  mean  proportional  between  2  and  8  is  4,  because  8X2= 
16 ;  £md  ^16=4. 

Proof.— 2  :  4  :  :  4 :  8;  2X8  =  4X4. 

Remark. — This  rule  is  evident  from  the  fact  that  a  mean  proportional  forma 
the  two  means  of  a  proportion.  (Sect-  XII.  Art.  2.  Obs.  7.)  But  the  product 
of  the  extremes  is  equal  to  the  product  of  the  means.  (Sect.  XII.  Art.  2.  Obs: 
9.)  Therefore,  the  square  root  of  th«  product  of  the  extremes  is  the  mean 
proportional. 

8.  What  is  a  mean  proportional  between  16  and  64? 

Ans.  32. 

9.  What  is  a  mean  proportional  between  68  and  612? 

Ans.  204. 

To  find  the  side  of  a  square  equal  in  area  to  a  given  surface. 

6  ft.  long-. 


— .«-  ..&J 

_ .._ 


9  ft. 

lon^. 

^ 

__J 

9X4=36  Jiq.  ft. 


6X6=36  sq.  it. 


We  perceive  from  the  diagrams  that  a  square  6  ft.  long,  and  6  ft. 
wide,  contains  6x6=36  sq.  ft.  We  also  perceive  that  a  rectangle 
9  ft.  long,  and  4  ft.  wide,  contains  9X4=36  sq.  ft.;  (Sect  IX.  Art. 
2.  Obs.  15.)  or  a  square  6  ft.  long  and  6  ft.  wide,  is  equal  in  area  to 
a  rectangle  9  ft.  long,  and  4  ft.  wide.  But  the  side  of  a  square  is 
formed  by  extracting  the  square  root  of  its  contents,  or  superficial 
area.  (Obs.  14.) 


^How  do  we  find  a  mean  proportional  betwe^m  two  numbers?     Show  why 
this  rule  is  correct.     How  do  we  fiud  the  side  of  a  square  equal  in  area  to  a 
given  surface?     Show  why  this  is  correct. 
15a 


322  COMMON    ARITHMETIC.  Scct.    XV 

Tlierefore,  if  we  extract  the  square  root  of  the  area  of  a  rectangle, 
'joe  shall  have  one  side  of  a  square  'of  equal  area,  as  the  result.  It  is 
ilso  evident,  from  Obs.  14,  that  the  square  root  of  the  area  of  any 
surface  is  one  side  of  a  square  surface  of  equal  area.     Hence — 

To  find  the  side  of  a  square  equal  in  area  to  a  given  surface  : 
Obs.  25.     Extract  the  square  root  of  the  given  area, 

10.  A  certain  field  is  48  rods  long,  and  12  rods  wide.  Re- 
quired— the  length  of  one  side  of  a  square  field  containing  the  same 
quantity  of  land?  Ans.  24  rds. 

11.  A  parallelogram  is  16  ft.  long,  and  9  ft.  wide.  What  is  one 
side  of  a  square  equal  in  area?  Ans.   12  ft. 

12.  A  circular  field  contains  2025  sq.  rds.  How  many  rods  on 
one  side  of  a  square  field  containing  the  same  quantity? 

Ans.  45. 

13.  A  certain  triangular  field  contains  10  acres,  128  sq.  rds.  Re- 
quired— one  side  of  a  square  field  containing  the  same  quantity.    "'. 

Ans.  41 .  569-)- rds. 

14.  If  a  certain  square  field  measures  30  rds.  on  a  side,  what 
will  be  the  length  of  one  side  of  a  square  field  containing  4  times  as 
much?  Ans.  60  rods. 

Operation. 
30 
30 

900  sq.  rds.  =  contents  of  the  given  field,  (Sect,  IX.  Art.  2. 
4  ^  [Obs.  15.) 

T '         ■       ' 

3600  sq.  rds.  =  contents  of  the  required  field. 


^3600  =  60  rds.,  one  side  of  the  required  field.   (Obs.  14.) 

15.  If  the  above  field  measured  28  rods  on  a  side,  what  is  the 
length  of  one  side  of  a  square  field  containing  9  times  as  much?  16 
times  as  much?     25  times  as  much?     64  times  as  much? 

Ans.  in  order.   84  rds,;   112  rds.;   140  rds.;  224  rds, 

\Q.  If  a  hall  measures  40  feet  in  length,  and  10  feet  in  width, 
what  is  the  length  of  one  side  of  a  square  room  9  times  as  large? 
\  as  large?    jg  as  large?         Ans.  in  order.  60  ft.;  10  ft.;   15  ft. 

17.  I  have  512  sq.  rds.  of  land  in  afield  which  is  twice  as  long 
as  it  is  wide.     Required  its  length  and  width? 

Suggestion. — If  the  field  is  divided  at  the  middle  of  the  sides,  it 


Art.  2.  EVOLUTION.  323 

will  form  iwo  equal  squares,  each  containing  512-J-2=  256  square 
rods.  Ans.  Length  32  rds.;  width  16  rds. 

18.  A  nurseryman  wishes  to  set  out  1280  frut  trees,  having  the 
length  of  his  orchard  5  times  its  width.  How  many  rows  must  he 
have,  and  how  many  trees  in  a  row? 

Ans.  16  rows,  and  80  trees  in  a  row.^ 

19.  A  certain  room  contains  300  sq.  ft.,  and  its  width  is  |  of  its 
lenoth.     Required  its  leno-th  and  width? 

Ans,  Length  20  ft.,  Width  15  ft. 

CIRCLES. 

Obs,  26.  A  Circle  is  a  portion  of 
space  enclosed  by  a  curved  line,  called  the 
Circumference,  ey^T^/^ar^  of  which  is  equally 
distant  from  aj^oint  mthin,  called  the  Cen- 
ter. 

A  straight  line  passing  through  the  center 
from  one  part  of  the  circumference  to  ano- 
ther is  called  the  Diameter. 

A  straight  line  from  the  center  to  any  part  of  the  circumference 
is  called  a  Radius. 

Thus,  in  the  above  figure,  the  point  A  is  the  Center;  the  line  B  E 
C  D  is  the  circumference;  the  line  B  A  C  is  the  Diameter;  and 
the  line  A  D  is  the  Radius. 

Remark  1.  It  will  be  perceived  that  A  B,  awd  A  C  are  radii,  and  that 
the  radius  is  half  the  diameter.  For  this  reason  all  radii  of  the  same  circle  are 
equal. 

2.  The  learner  must  not  confonh'dthe  circle  with  the  circumference.  The 
circle  is  the  surface  enclosed,  and  the  circumference  is  the  line  which  enclosesf 
it. 

Obs.  27,  The  areas  of  circles,  (and  also  of  all  oth^ surfaces,) 
are  to  each  other  as  the  squares  of  their  like  dimensions,  -,^ 

Note. — This  proposition  cannot  be  demonstrated  without  a',knowl«dgo  o 
geometry. 

To  find  the  dimensions  of  a  circle  which  shall  contain  £,  3,  4,  \, 
\,  &c.  times  as  much  as  a  given  circle: 

Obs.  28.  Square  the  diameter,  or  circumferen4:e  (whichever  is  giv- 
en i)  of  the  given  circle,  multif)ly  the  square  by  the  given  proportion. 


What  is  a  circle?  What  is  the  circumference?  The  diameter?  Tfie  radius? 
What  is  the  ratio  of  the  radius  to  the  diameter?  What  is  the  tWerenee  be- 
tween the  circle  and  the  circumference?  What  relation  have  rt^e  areas  of  cir- 
cles to  each  other?  How  do  we  find  the  dimensions  of  a  circb  which  shall  con- 
tain 2, 3,  4,  i,  h,  4,  &c.,  times  as  much  as  a  given  circle? 


324  COMMON    ARITHMETIC.  SeCt.    XVI 

and  extract  the  square  root  of  the  product;  the  result  will  be  the  similar 
dimensions  of  the  required  circle. 

20.  A  circle  measures  6  inclies  in  diameter.  Required  the  diam- 
eter of  one  4  times  as  large?  Ans.   12  inches. 

21.  The  circumference  of  a  circle  is  8  feet.  Required  the  cir- 
cumference of  one  9  times  as  large?  Ans.  24  feet. 

22.  The  circumference  of  a  circle  is  16  feet.  Required  the  cir- 
cumference of  one  27  times  as  large?  Ans  75  feet. 

23.  A  gentleman  has  two  circular  ponds  on  his  farm  ;  one  of 
which  IS  16  yards  in  diameter,  and  the  other  is  16  times  as  large. 
Required  the  diameter  of  the  larger?  Ans.  64  yds. 

24.  The  diameter  of  a  circle  is  1 6  feet.  Required  the  diameter 
of  a  circle  but  \  as  large?  Ans.  8  feet. 

25.  The  circumference  of  a  circle  is  64  feet.  Required  the  cir- 
cumference of  a  circle  but  y^  as  large?  Ans.  16  feet. 


SECTION  XVI. 
MENSURATION. 

Def.  Mensuration  teaches  the  art  of  finding  the  area  of  surfaces 
or  the  contents  of  solids.  The  area  of  a  surface  is  the  space  enclosed 
by  its  boundaries.     (Sect,  IX,  Art.  2,  Obs.  14.) 

Article  1.     Mensuration  of  Surfaces, 

Obs.   1.     When  we  speak  of  a  surface  v^t  mean  the  face  of  any 
thing.      A    Surface,  therefore  has  length   and   breadth^  but  not 
thickness.  *"   "^ 
'm- 

Case  1.     To  find  the  area  of  a  Square,  Rectangle y  <kc^ 

Obs.  2.  Multiply  together  its  length  and  breadth,  (Sect.  IX,  Art. 
2,  Obs.   15.) 

Hemark.  The  area  of  a  surface  is  expressed  by  Square  Measure;  as,  square 
feetf  square  rods,  square  miles,  &,c, 

f.   Ex.    1-  How  many  square  feet  in  the  floor  of  a  room  22  ft.  long 
md  lOii:    w^^e?  Ans.  352. 


What  is  Mens.  '"^*'°"^  What  is  the  area  of  a  surface?  When  we  speak  of 
a  surface  what  do  .  ^*^  mean?  What  then  has  a  surfae©?  How  dp  we  find  th& 
area  of  a  square,  rectu  '"*'^-  ^^' 


Art.    1.  ^     MENSURATION.  325 

2  A  square  field  measures  34  rds.    on  each  side.     Required,  its 
contents.  Ans.   1156  sq.  rds.  =  7  A.  36  sq.  rds. 

3.  A  certain  field  is  125  rods  long  and  65  rds.  wide.     Required, 
its  contents,  it  being  in  the  form  of  a  rectangle. 

Ans.  50  A.   125  sq.  rds. 

Cask  2.     To  find  the  area  of  a  Triangle. 

1.  How  many  square  feet  in  a  board  12  ft.  long,  and  4  ft.  wide  at 
one  end,  and  tapering  to  a  point  at  the  other?  Ans.  24. 

12  ft.  D         The  surface  of  the  board  would 

evidently  describe  a  triangle  (which 
we  will  suppose  to  be  right-angled,) 
like  the  triangle  A  B  C  in  the  dia- 
gram. Now  by  drawing  the  lines 
A  D  and  D  C,  parallel  to  B  C  and  A  B,  we  form  another  equal  and 
similar  triangel  ADC,  and  also  the  rectangle  A  B  C  D.  The 
length  and  width  of  this  rectangle  are  the  same  as  the  base  and  per- 
pendicular of  the  triangle  ABC;  that  is  12  feet  long,  and  4  feet  wide; 
and  its  area  is  12  X  4  =  48  sq.  ft.  (Obs  2.)  Therefore,  as  the 
rectangle  is  composed  of  two  equal  triangles,  the  area  of  each  must 
be  48  -i-  2  =  24  sq.  ft.     Hence — 

To  find  the  area  of  a  right  angled  triangle  : 

Obs.  3.     Multiply  together  the  base  and  perpendicidar,    and  take 
half  their  product. 

**  Remark  1.     ParaZZeZ  iincs  are  those  which  are  every  where  equally   distant 
from  each  other. 

2.  From  the  above  demonstration  we  perceive  that  every  Triangle  is  half 
a  Rectangle  havingthe  same  base  and  altitude. 

2.  What  is  the  area  of  a  triangle  the  base  of  which  is    40  feet, 
and  the  perpendicular  24  feet?  Ans.  480  sq.  ft. 

3.  How  many  teres   in  a  triangular   field   the   side   or  base  o^ 
which  is  124  rods,  and  the  end  or  perpendicular,  40  rods? 

Ans.   15  A.  80  sq.  rds. 

4.  How  many  acres  in  a  triangular  field,  the  base  of  which  is  150 
rods,  and  the  perpendicular  50  rds? 

Ans.  28  A.  20  squa;e  rods. 

"When  the  triangle  is  not  aright  angled  triangle,  and  we  have  the 
three  sides  given  we  proceed  as  follows: 

How  IS  the  area  of  a  surface  expressed?     How  do  we  find  the  area  of  a  right 
angled  triangle?    What  are  parallel  lines?     What  is  every  triangle? 


32G  COMMON    ARITHMETIC.  Scct.    XVI 

Obs.  4.  Add  together  the  three  sides  awd  take  half  their  sum. 
From  the  half  sum  take  the  three  sides  severally: 
Finally,  multiply  togetJierthe  half  sum  and  the  three  remainders,  and 
extract  the  square  root  of  the  product. 

Rbmark.     When  the  perpendicular  height  is  given  w©  work  by  Obs.  3. 

Note. — The  demonstration  of  this  rule  cannot  bo  understood"  without  a 
knowledge  of  geometry. 

5.  Required  the  area  of  a  triangle  the  sides  of  which  are  6,  7  and 
8  feet,  respectively?  Ans.  20.333+ sq.  ft. 

6.  Required  the  area  of  a  triangle  the  sides  of  which  are  24,  30,' 
and  36  feet  respectively? 

Ans.  357.176 +sq.  ft. 

7.  Required  the  area  of  a  triangle  the  sides  of  which  are  26,  28, 
and  30  feet,  respectively?  Ans.  336  sq,  ft. 

8.  How  many  sq.  yds.  of  plastering  are  there  in  a  triangle  whose 
sides  are  30,  40,  and  60  feet,  respectively?  Ans.  66f . 

Case  3.     To  find  the  area  of  a  Trapezoid. 

"~~"                  \  Obs.    5.      A  figure    having  tico 
\      sides  'parallel,  and  the  other  two  not  is 
\.  called  a  Trapezoid. 


1.  A  man  has  a  board  in  the  form  of  a  trapezoid;  the  length  of 
one  side  is  60  inches,  and  of  the  other  side  48  inches,  and  its  width 
is  24  inches.  *  Required  its  area?  Ans.  9  sq.  ft. 

r  60X48=  108  inches=the  length  of  both  sides. 

(  108-^2=54  inches=themean,  or  average  length.  (Sect. 
^  .     j       IX,  Art.  2,  Obs.  24.) 

uperation^  54X24=1296  sq.  in.=the  area  of  the  surface  of  the 
board.     (Obs.  2.) 

^  1296-r-t44=9  sq.  ft.     Hence— 

To  find  the  area  of  a  trapezoid: 

Obs.  6.     Multiply  half  the  sum  of  the  parallel  sides  by  the  width, 

2.  Required  the  area  of  a  trapezoid  whose  parallel  sides  are  64 
and  67  feet,  and  the  width,  39  feet.  Ans.  2359^  sq.  ft. 

3,  A  farmer  has  a  field  in  the  form  of  a  trapezoid  :  the  parallel 
sides  measure  40  and  50  rods,  and  the  width  measures  20  rods. — 
How  many  acres  does  it  contain?  Ans.  5  A.  100  sq.  rds. 

Case  4.     Circles  and  Polygons^ 


When    the  triangle  is  not   right  angled  how  do  we  find  is.  area?     What  is 
a  Trapezoid?     How  do  we  find  the  area  of  a  Trapezoid? 


Art.   1.  MENSURATION    OF    SURFACES.  327 

Obs.  7.     A  Polygon  is  a  jigure  hounded  by   straiglU  lines.     A 
Trianfjile  is  a  polygon  ;  also,  a  square. 

;  Problem  1.     To  find  the  area  of  a  regular  polygon  : 

Obs.  8.     Multiply  the  perimeter  by  half  the  perpendicular,  let  fall 
from  the  centre  to  the  middle  of  one  side. 


Demonstratio7i.-^jTi  the  diagram  we  diaw 
the  right  angled  triangle  O  S  A  within  the 
r-D  polygon  ABODE.  In  the  same  man- 
ner the  entire  polygon  may  be  divided  into 
right  angled  triangles.  But  the  area  of  the 
triangle  A  S  O  is  found  by  multiplying  the 
base  A  S  by  half  the  the  perpendicular  S 
])  Q         0;  (Obs.  3,)  and  the  area  of  the  other  tri- 

angles is  found  by  multiplping  their  bases  by  half  of  an  equal  per- 
pendicular. Therefore,  since  the  sum  of  their  bases  forms  the  per- 
imeter of  the  polygon,  the  rule  is  evident. 

Remark  1.  A  Regular  Polygon  is  one  in  which  each  corner  is  equally  dis- 
tant from  the  center.  The  middle  point  of  each  side  is  also  equally  distant 
from  the  center,  and  the  sides  are  all  of  equal  length. 


2.  A  polygon  of  five  sides  is  called  a  Pentagon;  of  six  sides,  a  Hexagon; 
of  seven  sides,  a  Heptagon-;  of  eight  sides,  an  Octagon;  of  nine  sides,  a  A''o7i- 
agon;  and  of  ten  sides  a  Decagon. 

1.  How  many  square  feet  in  the  floor  of  a  school  house  in  the 
form  of  an  octagon,  each  side  measuring  10  feet,  and  the  line  from 
the  center  to  the  middle  of  one  side  measuring  12  feet? 

Ans.  480  sq.  ft, 

2.  A  man  has  a  field  in  the  form  of  a  hexagon,  each  side  of 
which  is  60  rods,  and  the  distance  from  the  center  to  each  corner  is 
60  rods.     Required,  its  area?  AnS.  68  A.  8  sq.  rds.,  nearly. 

Problem  2.     To  find  the  area  of  irregular  Polygons  : 

Obs.  9.  Draiv  diagonals  dividing  the  polygon  into  triangles,  rect- 
angles, dr.  Then  find  the  area  of  each  of  these,  separately,  and 
add  these  several  areas  together. 

Remark.  A  Diagonal  is  a  lino  joining  two  angles  not  adjacent  lo  each 
other. 

What  is  a  Polygon?  Give  exampk*s.  How  do  we  find  the  area  of  regular 
polygon?  Demonstrate  tliis  rule?  W hut  is  a  regular  polygon?  What  is  a  pol- 
ygon of  five  sides  called?  Of  six  sides?  Of  seven  sides?  Of  eight  sides? 
Of  nine  sides?  Of  ten  sidea?  How  do  we  find  tht  area  of  an  irrrgular  poly- 
gon? 


328 


COMMON    ARITHMETIC. 


Sect.  XVI 


3:  There  is  a  certain  field  in  the 
form  of  the  adjoining  diagiam.  The 
side  A  B  measures  26  rds.;  the  side 
B  C  measures  28  rds.;  the  side  C  D 
measures  12  rds.;  the  side  D  E  mea- 
sures 20  rds.;  and  the  side"  E  A  mea- 
15  rds.;  diagonal  A  C  measures  30 
rds.;  and  the  line  F  G  measures  10 
rds.  How  many  acres  in  the  field? 
Ans.  3  A,  106  sq.  rds. 


Problem  3.     Circumscribed  and  Inscribed  Polygons. 


Obs.  10.  When  a  Polygon  is  drawn 
about  a  circle  so  that  the  middle  point  of 
each  side  touches  the  circumference,  it  is 
called  a  Circumscribed  Polygon. 

When  a  polygon  is  drawn  within  a  circle 
so  that  each  corner  shall  touch  the  circum- 
ference, it  is  called  an  Inscribed  Polygon. 

Thus,  in  the  diagram,  A  B  C  D  E  F  is 
a  circumscribed  polygon,  and  ab  c  d  ef  is 
an  inscribed  polygon.  Also,  in  the  same 
diagram,  the  circle  is  said  to  be  inscribed 
within  the  polygon  A  B  C  D  E  F,  and  to  be  circumscribed  about  the 
polygon  abc  d  ef. 

Remark  1.  The  term  Circumscribed  is  derived  from  two  Latin  words, 
Circum,  about,  and  Scriho,  to  write,  or  draw,  The  term  Inscribed  is  derived 
from  the  same  word  Scribo,  and  In,  within. 

2.  The  lines  which  bound  the  polygon  are  termed  the  Perimeter  of  the 
polygon. 

If  in  the  above  diagram  we  bisect,  or  divide  into  two  equal  parts 
each  side  of  each  polygon  it  is  evident  that  each  polygon  will  ap- 
pi  oach  nearer  the  circumference,  and  consequently  nearer  each  oth- 
er.    Therefore,  if  the  sides  of  each  polygon  be  bisected  indefinate- 


What  is  a  diagonal?  What  is  a  circumscribed  Polygon?  An  Inscribed  Pol- 
ygon? What  is  said  respecting  the  circle  in  the  diagram?  From  what  is  the 
term  circumscribed  derived?  What  does  it  mean?  From  what  is  the  term  in- 
frribed  derived?  What  does  it  mean?  What  is  the  Perimeter  of  a  Polygon? 
What  effect  does  it  have  upon  the  perimeter  of  a  circumscribed  and  inscribed 
polygon,  to  bisect  each  side  indefinately?  Where  will  the  perimeters  meet? 
Why? 


Art.    1.  MENSURATION    OF    SURFACES.  ^29 

ly,  the  perimeters  of  the  two  polygons  will  meet,  .  But  as  the  t»- 
scrihed polygon  cannot  fall  Avithout  the  circumference,  (for  then  it 
would  be  a  circumscribed  one,)  and  as  the  circumscribed  polygon 
cannot  fall  within  the  circumference,  (for  then  it  would  be  an  in- 
scribed one,)  they  must  meet  upon  the  circumference.     Hence — 

Obs.  11.  A  Circle  may  be  defined  as  a  polygon  uith  an  indefinite 
nu7)d)er  of  sides. 

Remark.  This  definition  is  not  strictly  correct,  as  the  side  of  a  polygon, 
and  the  circamference  of  a  circle  can  never  coincide  so  as  to  form  one  and 
the  same  line.     It  is  sufficiently  correct,  however,  for  all  practical  purposes. 

_r!PR0BLEM  4.     The  Quadrature  of  the  Circle. 

The  problem  is  not  only  difficult,  but  impossible  to  solve  exactly, 
although  we  can  approximate  very  nearly.  What  is  meant  by  it, 
is,  to  determine  the  area  of  a  circle,  the  diameter  of  which  is  equal 
to  the  side  of  a  given  square  ;  that  is,  to  find  the  area  of  a  circle  in- 
scribed in  a  square. 

The  first  step  is  to  find  the  ratio  of  the  circumferance  to  the  diameter. 
The  process  used  to  obtain  this  is  too  complicated  for  a  treatise  on 
Arithmetic;  therefore  we  will  merely  tell  how  it  is  done,  and  take  the 
result  as  correct  without  further  trouble.  The  learner  must,  how- 
ever, bear  in  mind,  that  the  exact  ratio  can  never  be  ascertained, 
neither  the  area  of  the  circle,  and  all  good  mathematicians  have 
long  since  abandoned  both  as  impossible. 

The  process  used,  is,  to  inscribe  and  circumscribe  a  circle  with 
regular  polygons,  and  then  to  bisect  these  until  the  sides  of  the  two 
polygons  and  the  circumference  of  the  circle  all  seem  to  unite  in 
one  common  line.  They  do  not  exactly  coincide,  (Obs.  11.  Rem.) 
but  come  sufficiently  near  to  render  accuracy  almost  absolutely  cer- 
tain. The  circumference  of  the  circle  must  always  be  between  the 
perimeters  of  the  two  polygons,  and  since  the  ratio  of  the  diameters 
to  the  perimeters  of  the  polygons  can  be  ascertained,  the  ratio  of  the 
diameter  to  the  circumference  of  the  circle  must  lie  between  these 
two  ratios. 

Archimedes  a  Grecian,  first  commenced  the  investigation  of  this 
problem,  and  by  increasing  the  number  of  sides  of  the  polygon  to 
32768,  he  found  the  ratio  to  be  between  3|^and  3i^.  3^^  =  \^,  or 
the  ratio  of  the  circumference  to  the  diameter  is  as  22  to  7.  Metius 


How  then  may  a  circle  be  defined?  Is  this  strictly  correct?  Why  not? 
What  is  meant  by  thf.  quadrature  of  the  circle?  Is  this  problem  capable  of 
being  solved  exactly?  What  is  the  first  «tep  in  solving  it  approximately? — 
What  is  the  process  used  to  find  this?  Can  the  perimeters  of  the  polygons, 
and  the  circumference  of  the  circle  ever  coincide?  Where  does  the  circumfer- 
ence of  the  circle  lie?  Where  then  must  the  ratio  of  the  diameter  to  the  cir- 
cumferaofe  be?    Why? 


330  COIVIMON     ARITHMETIC.  ScCt.    XVI 

a  German,  afterwards  found  the  ratio  to  be  as  355  to  113.  Van 
Ceulen,  a  Dutch  mathematician,  carried  the  process  much  farther, 
and  found  that  if  the  diameter  was  1,  the  circumference  would  be 
less  than  3.14159265358979323846264338327950289,  and  great- 
er than  3.14159265358979323846264338327950288.  Later  math- 
ematicians have  carried  the  process  still  farther,  and  have  ascertain- 
ed the  ratio  of  the  diameter  to  the  circumference  to  be  as  1  to  3.- 
141592653589793238462643383279502884197169399375705820- 
974944592307816406286208998628034825342117067982148086- 
5132823066470938446460955051822317253594081284802. 

The  ratio  then  may  safely  be  put  down  as  3.141592.  This  gives 
the  exact  ratio  to  5  decimal  places,  and  makingthe  error  as  small  as 
TooWoo--  For  all  practical  purposes,  3.1416  may  be  us;  d,  chang- 
ing the  9  in  the  fifth  order,  into  6  in  the  fourth  order  of  decimals. 
Hence — 

To  find  the  circumference  of  a  circle  when  the  diameter  is 
given  : 

Obs.  12.  Multiply  the  diameter  hy  3.1416;  or,  for  greater  accuracy, 
%  3.141592. 

1.  A  certain  wheel  is  36  in.  in  diameter.  Required  its  circum- 
ference. Ans.   9  ft.  5.0976  in. 

2.  If  the  diameter  of  a  circle  is  15  ft.,  what  is  the  circumference? 

Ans.  47.124  ft. 

3.  If  the  earth  is  7912  miles  in  diameter,  what  is  its  circumfer- 
ence? Ans.   24856.3392. 

To  find  the  diameter  of  a  circle  when  the  circumference  is  given. 

Obs.   13.     Divide  the  circumference  hy  ZA'ilQ  .      Or, 
Multiply  the  circumference  by  .31831. 

Remark.  Since  the  circumference  is  I  multiplied  bv  3.141^,  the  diameter 
isl  divided  by  3.1416.     1  divided  by  3.1416  equal  .31831. 

4.  If  the  circumference  of  a  circle  is  37.6992  ft.,  ^^llat  is  the  di- 
ameter? Ans.   12  ft. 

5.  If  the  circumference  of  a  circle  is  60  ft.,  what  is  its  diameter? 

Ans.    19.0985  ft. 
•     6.  If  the  circumference  of  a  circle  is  42  ft.,  6  in.,  what  is  its  di- 
ameter? Ans.    13  ft.  6.3376  in. 

7.  If  a  tree  measures  12  ft.  9  in.  around,  wliat  is  tlie  distance 
thouMi  it.  Ans.  4.0584  ft. 


To  what  may  this  ratio  bo  safely  put  down?  liow  small  is  Ihe  error  in  this 
cape?  What  number  is  used  for  practical  ])urpos;'S.  How  llien  ilo  vvc  find  the 
circumference  of  a  circle  when  the  diameter  i.s  given?  IIow  do  we  find  the 
diameter  when  the  circumference  is"iven? 


Art.     1.  MENSURATION    OF    SURFACES.  331 

Problem  5.     The  area  of  the  Circle, 

In  the  diagram  (Obs,  8.)  we  have  shown  the  right-angled  triangle 
a.  S  0,  the  area  of  which  is  equal  to  one-half  the  perpendicular  (0 
S)  multiplied  by  the  base  a  S.  (Obs.  3).  The  whole  polygon 
may  be  separated  in  triangles  in  the  same  manner,  and  the  area  of 
each  is  found  by  the  same  process.  Now  if  the  sides  of  the  poly- 
gon are  bisected  until  they  are  indefinitely  increased,  its  perimeter 
will  be  the  circumference  of  the  circle,  and  the  perpendicular  0  S 
will  be  the  radius  of  the  circle.  (Sect.  XV,  Art.  2,  Obs.  26.) 
Hence — 

Obs.  14*  The  area  of  a  circle  is  found  ly  mmltiplyiug  the  circum- 
ference by  half  the  radius* 

1.  If  a  circle  is  15  ft.  in  diameter,  and  47,124  feet  in  circumfer- 
ence how  many  square  feet  does  it  contain?  Ans.   176.716. 

The  learner  will  remember  that  the  radius  is  half  the    diameter 
Sect.  XV,  Art.  2,  Obs.  26,  Rem.  1.) 

2.  If  the  diameter  of  a  circle  is  25  ft.,  and  the  circumference 
78.64  ft.,  what  is  its  area?  Ans.  490  sq.  ft.,  126  sq.  in. 

As  the  radius  is  half  the  diameter,  half  the  radius  is  \  of  the 
diameter,  the  area  of  a  circle  is  equal  to  the  product  of  the  circum- 
ference into  \  the  diameter;  or  which  is  the  same  thing  \  of  the 
product  of  the  circumference  and  diameter.  But  the  circumference 
is  equal  to  the  diameter  multiplied  by  3.1416;  therefore,  (Obs.  12.) 
the  circumference  multiplied  by  the  diameter  is  the  same  as  3.1416 
times  the  diameter  into  itself,  or  3.1416  times  the  square  of  the  di- 
ameter ;  and  the  area  of  the  circle  is  \  of  this,  or  3.1416  -r-  4  = 
.7864  times  the  square  of  the  diameter.     Hence — 

a.  The  area  of  a  circle  may  he  ftund  ly  muliijAying  the  square  of 
the  diameter  hy  .7864. 

Remark.  The  learner  will  observe  from  this  that  when  the  diameter  of  the 
circle  i>  1,  the  area  is  .7654. 

How  do  we  find  the  area  of  the  circle  when  the  circumference  and  diameter  - 
are  both  given?     Demonstrate  this  rule.     When  the  diameter  only  is  given  how 
do  we  find  the  area?     Demonstrate  this  rule.     When  the  circumference  only 
is  given  how  do  we  find  the  area?     Demonstrate    thi«   rule?     When   we  have 
given  the  area  how  do  we  find  the  dimensious? 


*In  this  cafe  as  the  rircumfereiiro  is  tlie  I'.eriineter  of  a  polv^nn  witli  an  indefiiiie  rmiibcr 
of  side.;,  the  fircle  is  composed  of  an  iiKleliiiite  buwI  ov  of  triaii^ics,  of  which  !he  ar.-s.  of  carli 
is  fouTi'l  by  Ol's.  3,  and  the  sum  of  all  these  areas  is  rqua!  to  the  a--(a  of  the  citric.  !?ect. 
IV,  Art.^  Obs.  4.  Rem.  2).  Or,  as  the  per|>en(|jcu!ars  to  all  tliese  triaiijiles  are  t-qiia!,  the 
siiii!  of'il^ir  bases,  (or circumferences)  multipiicd  by  ha'f  the  ])ericjidi(ular  (or  radius;  iiiusl 
"ivethe  area. 


332  COMMON    ARITHMETIC.  ScCt.    XYI 

3.  Required — the  area  of  a  circle  tlie  diameter  of  which  is  60  ft, 

Ans.   1963|sq.  ft. 

When  the  circumference  is  1,  the  diameter  is  .31831,  (Obs,  13, 
Rem.)  and  the  area  .31831^  X  .7854.  But  .31831^  X  .-7854=:  .08 
nearly.     Hence — 

h.  The  area  of  a  circle  may  be  found  by  multiplying  the  square  of 
the  circumference  by  .08;  or,   for  greater  acuracy  by  .0795, 

4.  If  the  circumference  of  a  circle  is  1 5  ft.  what  is  its  area?  ; 

Ans.   18  sq.  ft, 

5.  If  the  circumferance  is  28  ft,  what  is  the  area? 

Ans.  62.72  sq.  ft. 

6.  The  area  of  a  circle  is  176.715  sq.  ft.,  what  is  its  diameter 
and  circumferance? 

Ans.  Dia.  15  ft.,  Cir,  47.124  ft. 

OrPration  \  176.715-^  ,7854  =  225  ;  ^225  ,  =  15  ft.  diameter. 
^  ^  15  X  3,1416  =  47.124  ft.  circumference. 

Remark.  This  question  is  txactly  the  reverse  of  those  under  Obs.  14,  a. 
Therefore,  the  rule  is  evident.     Hence — 

When  we  have  the  area  of  a  circle  given  to  find  the  dimensions: 

Obs.  15.  Divide  the  area  by  .7854,  and  extract  the  square  root  of 
the  quotient  for  the  diameter.  The  circumference  is  then  found  by  Obs, 
12. 

7.  A  square  field  contains  490,875  sq.  rds.  Required — the  di- 
ameter of  a  circular  field  containing  the  same  quantity.  The  cir- 
cumference. 

Ans.  Dia.  25  rds..  Circle  78.54  rds. 

8.  How  long  must  a  halter  be,  which  being  fastened  to  a  post  at 
the  center,  will  just  allow  a  horse  to  feed  on  half  an  acre  of  ground? 

Ans.  83  ft.  3.1476  in. 

Problem  6.     To  find  the  Area  of  a  Square  inscribed  in  a  Circle. 

The  rule  is  deduced  from  the  following  proposition  : 

1.  Tlie  area  of  a  square  inscribed  within  a  circle  is  one-half  the 
area  of  a  square  circumscribed  about  the  same  circle. 


From  what  principle  is  the  Rule  for  finding  the  area  of  a  square  inscribed  in 
a  circle  deduced?  Demonstrate  this  principle.  To  what  is  theareaef  an  in- 
scribed square  equal? 


Art.  1. 


MENSURATION    OF    SURFACES. 


U9 


B 


C 


H 


t^  1^ 

iX                                 \ 

^ 

D 


Deynonstratlon. — Let  A  B  C  D  be 
the  inscribed,  and  E  F  Gr  H  the  cir- 
cumscribed square.  It  will  be  per- 
ceived that  the  angles  of  the  inscri- 
bed square  are  located  at  the  middle 
of  the  sides  of  the  circumscribed 
square.  Draw  the  diagonals  BOD 
and  C  0  A.  These  divide  the  cir- 
cumscriiied  square  into  four  equal 
smaller  squares.     The  sides  of  the 

inscribed  square  divide  each  of  these  smaller  squares  into  two  equal 
and  right  angled  triangles.  (Obs.  3,  Rem.)  Therefore,  since  both 
of  these  triangles  are  included  in  the  circumscribed  square,  and  but 
one  of  them  in  the  inscribed  square,  the  circumscribed  square 
contains  eight  triangles,  and  the  inscribed  square  but  four — hence, 
the  proposition  is  correct. 

Now  the  side  of  the  circumscribed  square  is  equal  to  the  diam- 
eter of  the  circle,  and  its  area  is  the  square  of  this  diameter. — 
(Obs.  2.)     Hence — 

Obs.  16.  The  area  of  a  square  inscribed  within  a  circle  is  one  half 
the  square  of  the  diameter  of  the  same  circle. 

a.  The  area  of  an  inscribed  square  is  also  equal  to  the  diameter  of 
the  circumscribed  circle  multiplied  into  one  half  of  itself. 

This  is  evident  from  an  examination  of  the  above  diagram.  We 
perceive  that  the  inscribed  square  is  composed  of  the  two  equal  tri- 
angles, A  B  C  and  A  p  C.  (Obs.  3,  Rem.  2.)  Now  the  area  of 
each  of  these  triangles  is  equal  to  the  hypothenuse  A  C,  multiplied 
by  half  the  perpendicular  height  B  0,  or  0  D,  (Obs.  3.)  But  ^  of 
B  0  +  i  of  0  D  is  equal  to  either  B  0  or  0  D,  since  B  0  and"  0 
D  are  equal.  Hence — the  area  of  the  inscribed  square  is  equal  to 
A  C  (the  diameter  of  the  circumscribed  circle)  multiplied  by  B  0, 
Or  0  D,  (half,  this  d'ameter.) 

1.  Required —  the  area  of  a  square  inscribed  in  a  circle  12  feet 
in  diameter:  Ans.  72  sq.  ft. 

Solution.— \2  X  12  =  144;  144  -^-  2  =  72.  Or,  12X6  = 
72. 

2.  Required— the   cost  of  a  log  when   squared,  at   $0.40  per 

To  what  else  is  it  equal?  Why?  How  do  we  find  the  side  of  an  inscribed 
square?  Why  is  this  correct?  When  the  diameter  of  the  circle  only  is  giv- 
en how  do  we  find  the  side  of  the  inscribed  square?  Demonstrrate  this  rule. 
When  the -circumference  only  is  given  how  do  we  find  the  area?  Demon- 
strate  this  rUle?     To  what  may  this  principle  be  applied? 


334  COMMON    ARITHMETIC.      ir  Scct,    XVI 

square  foot,  which  measures  3  ft.  in  diameter,  and  22  ft.  in  length. 

Ans.  $39.60. 

Suggestion. — Find  the  size  which  the  log  will  square,  and  multi- 
ply this  by  the  length  for  the  number  of  sq.  ft. 

3.  How  much  would  a  log  cost  at  $4.50  per  100  ft.,  which  is  4^ 
ft.  in  diameter,  and  48  ft.  inkngth?  Ans.  $21.87. 

2.  How  much  would  a  log  cost  which  is  4i-  ft.  in  diameter  and  64 
ft.  long,  at  $50  per  M.?  Ans,  $32.40. 

5.  How  much  would  a  log  cost  which  is  b\  ft.  in  dia.,  and  80  ft. 
long  at  $75  per  M.?  Ans.  $82.68|. 

Problem — To  tind  the  side  of  a  Square  inscribed  in  a  Circle, 

The  side  of  a  square  is  equal  to  the  square  root  of  its  are  a 
(Sect.  XY,  Art.  2,  Obs.  25).     Hence— 

To  find  the  side  of  an  inscribed  square  : 

Obs.  17.  Extract  the  square  root  of  its  area  found  according  to 
Obs,   16. 

From  examining  the  diagram  for  illustrating  the  preceding  case, 
we  perceive  that  the  inscribed  square  consists  of  four  equal  right- 
angled  triangles,  and  that  each  of  these  triangles  has  two  of  its 
sides  equal,  since  these  sides  are  radii  of  the  same  circle  (Sect.  XY, 
Art,  2,  Obs.  26,  Rem.  1).  The  hypothenuse  of  these  triangles  then 
is  equal  to  one  of  these  sides  (or  radii)  multiplied  by  1.4142.  (Sect. 
XV,  Art.  2,  Obs.  23,  Rem  1).  But  this  hypothenuse  is  the  side 
of  the  inscribed  square,  and  twice  the  other  side  of  the  triangle, 
(radius)  is  the  diameter  of  the  circumscribed  circle.  (Sect.  XV, 
Art.  2,  Obs.  26,  Rem.  1).  Therefore,  this  diameter  multiplied  by 
.7071  ( 1,4142 -f- 2)  must  be  the  side  of  the  inscribed  square. — 
(Sect.  VI,  Artl,  Obs.  30.) 

Again,  when  the  circumference  is  1  the  diameter  is  .31831,  (Obs, 
13,  Rem  )  and  when  th  3  diameter  is  1  the  side  of  the  inscribed  square 
is  .7071;  theretore  when  the  circumferance  is  1,  the  side  of  the  in- 
scribed square  is  ,31831  X  •7071  =:^  .2251  nearly.     Hence— 

To  find  the  side  of  an  inscribed  square  : 

a.  Multijyhj  the  diameter  hy  .7071,  or  the  circumference  hy  .2251, 

1 .  How  large  a  square  can  be  hewn  from  a  log  2  ft.  in  diame- 
ter. 

How  many  dimensions  are  used  in  measuring  lumber?  What  are  they? — 
What  is  the  thickness  considered?  How  is  lumber  more  than  an  inch  iu  thick- 
nesscounted?  When  the  waste  by  sawing  is  considered  how  do  we  proceed? 
When  the  lumber  ia  less  than  an  inch  in  thickness  how  do  we  proceed?  How 
do  we  find  the  dimensions  of  a  circumscribed  circle  when  the  side  of  the  in- 
scribed square  is  given?     Demonstrate  this  rule. 


Art.    1 .  MENSURATION    (  F    SURFACES.  335 

Solution.-^2''=4;  4-f-2  =  2;  ^2=  1.4142.  Or,  .7071  X  2= 
1.4142  as  before.  Ans.  1.4142  ft, 

2.  How  many  inches  will  a  log  square  that  is  120  inches  circum- 
ference? Ans.  27.012  in. 

Soluiion.— 120  X  .2251  =  27.0120. 

3.  I  wish  a  sill  18  inches  square,  and  find  a  tree  25  in.  in  diame- 
ter.    Will  it  answer  my  purpose  or  not? 

Ans.  It  will  not  as  it  will  square  but  17.G7  J  in. 

4.  I  find  another  tree  measuring  80  in.  in  circumlerence.  Will 
this  answer  my  purpose?       Ans.  It  will,  as  it  squares  18.008  in. 

This  principle  may  be  applied  to  finding  the  quantity  of  lumber  a 
log  will  make. 

5.  A  log  measures  24  ft.  in  length,  and  15  inches  in  diameter 
how  many  feet  of  inch  lumber  will  it  make,  allowing  no  waste  for 
sawing? 

Solution. — 15  X  -7071  =  10.6065  inches  that  the  log  will  square. 
Then  24  ft.  =  288  in.;  10.6065^=  area  of  the  end,  and  10.6065^ 
X  288  =  square  contents  in  inches,  which  divided  by  144  =  214.- 
389  +  sq.  ft.  Ans. 

The  operation  may  be  shortened  somewhat  by  canceling. 

Remark  1.  In  measuring  lumber  but  two  dimensions  are  generally  usee', 
viz:  length  and  width.  The  thickness  is  considered  a  unit  1  inch  being  the 
standard.  Lumber  more  than  an  inch  in  thickness,  is  counted  an  additional 
thickness  for  every  additional  inch.  When  the  waste  by  sawing  is  cousidered, 
it  should  be  deducted  from  the  result. 

6.  A  log  measures  25  feet  in  length,  and  12  inches  in  diameter. 
How  much  two  inch  lumber  will  it  make  allowing  it  to  lose  \  of  an 
inch  in  sawing?  Ans.   132.3225  sq.  ft. 

Solution. — We  find  the  contents  of  the  log,  as  in  the  last  exam- 
ple, to  be  150  sq.  ft,  nearly.  The  log  squares  8.4852  in.,  and  as 
the  lumber  is  2  inches  thick,  the  saw  passes  through  four  times, 
which  makes  1  inch  waste.  This  1  inch  measures  8.4852  in.  in 
width,  and  300  in,  (25  ft.)  in  length  which  makes  17.6775  sq,  ft. 
loss  by  sawing,  and  150—17.6775=  132,3225  sq.  ft  of  lumber. 

Remark  2.  Had  the  width  of  the  lumber  been  given,  and  the  log  been 
sawed  both  ways  across  the  end,  an  additional  waste  wouidhave  accrued  from 
sawing  which  would  have  been  deducted.  When  the  lumber  is  less  than  an 
inch  in  thickness  the  calculations  for  the  quantity  are  made  by  the  inch  as  usual, 
although  the  allowance  for  waste  in  sawing  must  be  made  according  to  the 
thickness  of  the  lumber  sawed. 


336  COMMON    ARITHMETIC.      ^  Scct.    XVI 

7,  A  log  measuring  2  ft.  8  in.  in  diameter  and  24  ft.  in  length,  is 
to  be  sawed  into  lumber,  6  inches  wide,  and  1  inch  thick.  Eequired 
the  cost  of  this  lumber  at  '^1.75  per  C,  the  waste  being  y^  of  an 
inch  for  sawing.  Ans,  -$5.17,  nearly. 

8,  Twelve  logs,  the  average  diameter  of  which  is  2  ft.  6  in.,  and 
the  lengths,  12,  15,  10,  22.  14,  16,  20  18,  15,  14,  24  and  20 
ft.  respectively  are  to  be  sawed  into  lumber  li-  inches  in  thickness, 
and  6  inches  in  width.  Required  the  cost  of  the  lumber  at  $22.50 
per  M,  allowing  y  g  of  an  inch  waste  for  sawing. 

Ans.  $142,392+. 

9,  How  much  half  inch  lumber,  4  inches  in  width,  can  be  sawed 
from  a  log  3  ft.  in  diameter,  and  25  ft.  long,  allowing  \  of  an  inch 
waste  for  sawing?  Ans.  589.488-|-sq.  ft. 

Problem  8.  To  find  the  Circumference  or  Diameter  of  the  Cir- 
cumscribed Circle  when  the.  side  of  the  Inscribed  Square  is  given. 

By  examining  the  diagram  for  finding  the  area  of  the  inscribed 
square,  we  perceive  that  the  diagonal  ot  this  square,  is  the  hypoth- 
enuse  of  a  right-angled  and  equal  sided  triangle,  the  legs  of  which 
are  the  sides  of  the  inscribed  square.  But  this  diagonal  is  the  di- 
ameter of  the  circumscribed  circie,and  is  equal  to  either  of  the  sides  of 
the  inscribed  square  multiplied  by  1.4142.  (Sect.  XV.  Art.  2, 
Obs.  23,  Rem.  1). 

Again,  the  circumference  is  equal  to  the  diameter  multiplied  by 
3.1416.  (Obs,  12.)  But  the  diameter  is  equal  to  the  side  of  the 
inscribed  square  multiplied  by  1.4142.  therefore  the  circ.  is  equal  to 
the  side  of  the  inscribed  square  X  1.4142  X  3.1416.  1.4142  X  3.- 
1416  =  4.443  nearly;  therefore  the  circumference  is  equal  to  the 
side  of  the  insciibed  square  multiplied  by  4.443.     Hence — 

To  find  the  dimensions  of  circle  when  the  side  of  an  inscribed 
square  is  given: 

Obs,  18.  Multiply  the  side  of  the  inscribed  square  1)1/  1.4142  for 
the  diameter y  or,  by  4.443  for  the  circumference. 

1 .  How  large  must  a  tree  be  in  diameter  and  circumference  to 
square  10  inches? 

Solution. — 10  X  1.4142  =  14.142  in.  dia,  and  10  X  4.443=  44. 
43  in.  Circ,  Ans. 

How  do  we  find  the  side  of  a  square  of  equal  area  to  a  given  circle?  De- 
monstrate this  rule.  How  do  we  find  the  dimensions  of  a  circle  of  equal  area 
to  a  given  square?     Demonstrate  this  rule. 


Art.    2.  MENSURATION    OF    SOLIDS.  337 

2.  I  wish  a  stick  of  timber  12  inches  square,  and  find  a  tree    16 
inches  in  diameter.     Will  it  answer  my  purpose  or  not. 

Ans.  It  will  not,  as  to  square  12  in.  it  must  be  16.9704  inches  in 
diameter. 

3.  How  large  must  a  tree  be  in  circumference  to  square  1 2  inches? 

Ans.  53.316  inches. 

4.  How  large  must  a  tree  be  in  diameter  to  square  15  inches? 

Ans.  21,213  inches. 

We  learn  (Obs.  14.  a.  Rem,)     that  if  the  diameter  of  a  circle  is 
1,  the  area  is  .7854  ;  therefore,  the  side  of  a  square  of  equal  area 
is  ^.7854  =  .8862;  and  .8862-r-  3.1416  =  .2821  nearly.  Hence— 
To  find  the  side  of  a  square  equal  in  area  to  a  given  circle: 

Obs.  19.  Multiply  the  diameter  of  the  circle  by  .8862;  or  the  cir- 
cumference by  .2821. 

6.  A  circle  is  12  ft.  in  diameter.  Required  the  side  of  a  square 
of  equal  area.  Ans.   10.6344  ft. 

6.  A  circle  is  40  ft.  in  circumference.  Required — the  side  of  a 
square  of  equal  area.  Ans.  11.284  ft, 

Since  the  diameter  of  a  circle  multiplied  by  .8862,  or  the  circum- 
ference multiplied  by  .2821,  will  give  the  side  of  a  square  of  equal 
area,  it  is  evident  that  the  side  of  a  square  divided  by  .8862  will 
give  the  diameter, or  by  .2821  will  give  the  circumference  of  a  circle  of 
equal  area.  Then  assuming  unity,  or  1  as  the  standard,  1  -r-  .8862 
=  1.128  as  the  diameter,  and  1  -7-  .2821  =  3.545  nearly,  as  the  cir- 
cumference of  a  circle,  the  area  of  which  is  1,  or  equal  to  the  area 
of  a  square  the  side  of  which  is  1 .     Hence — 

'To  find  the  dimensions  of  a  circle,  the  area  of  which  shall  be 
equal  to  the  area  of  a  given  square. 

Obs.  20.  Multiply  the  side  of  a  given  square  by  1. 1 28 /br  the  di- 
ameter; or^  by  3.545  for  the  circumference. 

7.  The  side  of  a  square  measures  150  feet.  Required — the  di- 
ameter and  circumferenc  3  of  a  circle  of  equal  area. 

Ans.  Dia.  169.2  ft.  Circ.  631.75  fl. 

Article  2.     Mensuration  of  Solids, 

Obs.  1.  In  Mensuration  of  solids,  two  things  are  to  be  consid- 
ered: 

1st,     T  he  Mensuration  of  their  surfaces;  9,nd 


How  many  things  are  tu  be  considered  iji  the  mensuratimi  of  solldi? 
IB 


338 


COMMON    ARITHMETIC. 


Sect.  XVI 


2nd.     The  mensuraiion  of  their  Solidities. 
A  Solid  has  length,  breadth,  and  thickness. 

Remark.     It  is aa  established  theorem  iu  G^omstry,  that  all  solid  bo  lies  are 
to  each  other  as  the  cubes  of  their  like  dinic'raions. 

1.  If  a  ball  weighing'  6  lbs.,  be  4  inches  in  diameter,  what  is  the 
diameter  of  a  ball  of  the  same  metal  weighing  48  lbs. 

Ans.  8  mches. 

Operation.— Q  lbs.  :  48  lbs.  :  \  64  (4^)  ;  612  =^512  =  8. 

2.  If  a  ball  8  inches  in  diameter  weighs  48  lbs,,  what  is  the  weight 
of  a  ball  4  inches  in  diameter?  Ans.  6  lbs. 

3.  If  a  cubical  vessel  is  8  inches  in  length,  what  is  the  side  of  a  cu- 
bical vessel  that  shall  contain  27  times  as^uch? 

Ans.  24  inches. 
Case  1.     Prisms. 


Obs.  2.  A  Prism  is  a  solid  having  its  ends,  or  bases 
equal  and  parallel.  It  is  sail  to  be  Vriangalar,  Qiad- 
rangular,  (kc,  according  as  its  sides  are  triangles, 
squares,  <fec. 


To  find  the  area  of  the  surface  of  a  prism  : 

Obs.  3.  Multiply  the  perimeter  of  the  lase  by  its  oltHude,  or  heighth, 
and  to  theprodmi  add  the  area  of  the  bases  when  the  entire  surface  is 
required. 

Demonstration. — The  sides  of  the  prism  are  parallelograms,  (Obs. 
2.)  and  the  area  of  each  of  these  parallelograms  is  found  by  multi- 
plying together  its  length  and  v/idth  ;  (Art.  1,  Obs.  2).  But  as  their 
lengths  are  equ  1,  their  area  is  equal  to  the  sum  of  their  width,  or 
bases  multiplied  by  their  length,  or  altitnde;  hence  the  above  rale 
is  evident 

Ex.  1.  What  is  the  entire  surface  of  a  triangular  prism,  whose 
sides  are  3,  4,  and  5  ieet,  and  whose  altitude  is  6  f  et: 

Ans.  84  sq.  ft. 

What  are  they?  What  has  a  solid?  What  relations  have  solid  bodie.s  to 
each  other?  What  is  a  prism'?  Whf-n  is  a  pnsni  triaiigular,  oua  irangular,  vlk,c.? 
How  do  we  find  the  surface  uf  a  prism?    Demonstrate  this  rul«? 


Art.  2. 


MENSURATION    OF    SOLIDS. 


33» 


2.  What  is  the  entire  surface  of  a  triangular  prism  the  sides  of 
which  are  8,  10,  and  12  feet,  and  its  altitude  9  feet? 

Ans.  349.36-j-sq.  ft. 

3.  What  is  the  convex  surface  of  a  quadrangular  prism,  each  side 
of  which  is  5  feet,  and  its  altitude  6  feet?  Ans.   120  sq.  ft. 

To  find  the  solid  contents  of  a  prism  : 

Obs.  4.     Multiply  the  area  of  its  base  by  its  altitude y  or  heighth. 

Demonstration. — The  area  of  the  base  comprises  the  two  dimen- 
sions, breadth  and  thickness,  and  is  expressed  by  square  measure. 
(Art  I,  Obs.  2.  Rem.)  But  the  altitude  is  the  length  of  the  prism; 
and  square  measure  multiplied  by  linear  measure  gives  solid  or  cu- 
bic measure.  (Sect.  XIV.  Rem,  1,,  under  the  Rule.)  Hence,  the 
rule  is  evident. 

4.  Required,  the  solid  contents  of  a  triangular  prism,  whose 
length  is  12  mches,  and  each  side  of  its  base  3  inches? 

'3_(_3_|-3=,9;  9-7-2=4.5=  the  half  sum  of  the  sides 

of  the  base. 
4.5 — 3=1 .5;  the  three  remainders  are  equal;  since 

the  sides  are  equal. 
4.5X1.6X1.5X1.5=15.1875;   ^15. 1875=  area 

of  base.   (Art.  1.  Obs.) 
^15.1875  X  12  =  contents   of  the    prism.       12  = 

V144. 
715.1875X    ^144  =  72187  =  46.7653+   sq. 

in.  Ans. 

5.  Required,  the  solid  contents  of  a  quadrangular  prism  36 
inches  in  length,  and  each  side  of  its  base  8  inches? 

Ans.   1  s.  ft.  576  s.  in. 

6.  Required,  the  solid  contents  of  a  triangular  prism,  whose  sides 
are  4,  6,  and  8  feet,  and  its  length  16  feet? 

Ans.   174.2812+ s.  ft. 

Case  2. — Cylinders. 


Obs.  5.  A  rotind  prism  having 
equal  circles  for  its  ends,  is  called  a 
Cylinder. 


To  find  the  surface  of  a  Cylinder  : 

Obs.  6.     Multiply  the  circumference  of  its  end  by  its  leng'h. 

How  do  we  find  the  solidity  of  a  priKiii?     Dmonsiraif- this  Rule.     What  is  a 
Cylinder?     How  do  we  find  the  surface  of  a  C5-IiDder? 


Operation. 


^ 


340  COMMON    ARITHMETIC.  ScCt    XVI 

Demonstration, — The  surface  of  any  prism  whose  base  is  a  poly- 
gon, is  found  by  multiplying  the  perimeter  of  its  base  by  its  alti- 
tude. (Obs.  3.)  Now  if  the  sides  of  a  base  are  bi  ected  indefi- 
nitely, the  perimeter  will  become  a  circle,  and  therefore  the  prism 
will  become  a  cylinder.  Hence,  the  rule  is  evident  from  Art.  1. 
Obs.  9. 

Remark. — The  above  rule  only  gives  the  convex  surface.  When  the  en/ ire 
surface  is  required,  the  area  of  tlie  banes  must  be  added  to  the  result. 

1 .  A  cylinder  measures  6  inches  in  circumference,  and  1 8  inches 
in  length.     Required,  its  convex  surface.  Ans.   108  sq.  in, 

2.  A  cylinder  is  60  inches  in  diameter,  and  15  feet  in  length. 
Required,  its  entire  surface.  Ans.  274.89  sq,  ft. 

To  find  the  solidity  of  a  cylinder  : 

Obs.  7.     Multiply  the  area  of  its  base  hy  its  length. 

Demonstration. — The  solidity  of  any  prism  whose  base  is  a  poly- 
gon, is  found  by  multiplying  the  area  of  its  base  by  its  altitude. 
(Obs.  4.)  Now  if  the  sides  of  the  prism  are  bisected  indefinitely, 
the  prism  will  become  a  cylinder.  (Art.  1.  Obs,  9.)  Hence,  the 
rule  is  evident. 

3.  Required,  the  solid  contents  of  a  cylmder  12  feet  in  length, 
and  5  feet  in  diameter?  Ans.  235.62  s.  ft. 

4.  How  many  solid  inches  in  a  common,  or  Winchester  bushel, 
the  diameter  of  which  is  is  18^  inches,  and  the  depth  8  inches. 

Ans,  2150.4252  s.  in. 

Case  3. — Contents  of  Boilers. 

Obs.  8,  A  Boiler  may  he  regarded  as  a  cylinder,  with  several 
smaller  cylinders  (flues)  udihin  it. 

A  cubic  foot  contains  1728  solid  inches  ;  a  wine  gallon  contains 
231  cubic  inches,  and  a  beer  gallon  contains  282  cubic  inches. 
Therefore  a  solid  foot  contains  1 728 -f- 23 1  =  7  .  48  wine  gallons  ; 
and  1728  —  282  =  6.127  beer  gallons.  1 728  X -7854=1 1357. 171 2 
cubic  inches  in  a  cylindric  foot;  and  1357  .  1712 -j- 231  =  5.875 
wine  gallons,  and  13-57.  1712 -f- 282  =  4  .  812  beer  gallons  in  a 
cylindric  foot,  5  .  875  =  f|>Ji  =  V  ;  and  4.812  =  J  J^ol  =  %\ 
Hence — 

To  find  the  contents  of  Boilers : 

Obs,    9.     Muiliply  the  square  of  the  diameter  hy  the  length  and 

Demonstrate  this  rule.  What  surface  does  this  rule  give?  When  the  en- 
tire surface  Is  required,  how  do  we  proceed!  How  do  we  find  the  soli  illy  of  a 
cylinder?  Demonstrate  this  rule.  As  what  may  a  boiler  be  regarded?  How 
do  we  find  the  cbntents  of  a  bb'ilerT     Demonerttate  this  rule. 


Art.    2.  MENSURATION    OF    SOLIDS.  341 

47,  and  divide  the  result  hy  8.  The  result  will  he  in  ''wine  gallons- 
For  beer  gallons  use  24  and  5,  instead  of  47  and  8.  Find  the  conn 
tents  of  the  fines  in  the  same  way^  which  subtract  frcm  the  entire 
contents. 

Remark  1. — The  dimensions  must  be  taken  in  feet  by  this  Rule 

I .  How  many  wine  gallons  in  a  boiler  30  inches  in  diameter,  and 
25  feet  long,  having  two  flues,  each  9  inches  in  diameter? 

Operation. 


30  iQ.=2^=|  'ft;  (1)^  X25X47-v-8=917.96875  entire  contents. 
9  in.=|ft.;  (J)^X25X47X2-r-8    =165.234375  flues. 


Ans.  762.734375  gallons. 
We  multiply  by  2  because  there  are  are  2  flues. 

2.  Required,  the  contents,  in  wine  and  beer  gallons,  of  a  boiler 
36  in.  in  diameter,  and  45  ft.  long,  having  3  flues,  each  8  inches  in 
diameter?  Ans.  2026.875  wine  gallons  ;  1656  beer  gallons. 

Remark  2.— The  contents  of  circuJar  cisterns  may  also  be  found  by  the 
above  rule.  V\  hen  the  cistern  is  square,  rectangular,  &c.,  multiply  its  solid 
contents  by  7.48  for  wine  gallons,  and  6.127  for  beer  gallons.  In  this  case  we 
do  not  use  the  numbers  47  and  8,  or  24  and  5.  Notice  that  the  dimensions 
must  all  be  taken  in  feet. 

Note. — The  wine  measure  of  231  cubic  inches  to  the  gallon,  is  generally 
used  as  the  standard  in  the  United  States,  and  is  so  understood  in  the  following 
examples,  unless  otherwise  mentioned. 

3.  RequTed,  the  contents,  in  wine  and  beer  gallons,  of  a  circular 
cistern,  6  It.  8  in.  in  diameter,  and  8  ft.  deep. 

Ans.  2088J  win  '  gallons;  1706^  beer  gallons. 

4.  Required,  the  cont  nts,  in  wine  and  beer  gallons,  of  a  rectan- 
gular cistern  that  contains  300  solid  feet? 

Ans.  2244  wine  gallons;  1838.1  beer  gallons. 

5.  A  rectangular  cistern  is  to  be  made  containing  1400  gallons  ; 
its  length  and  breadth  are  8  feet  each.     Required,  its  depth? 

Ans,  3  ft.  nearly. 

Suggestion. — Find  the  contents  of  the  cistern  by  multiplying  1400 
by  231.  We  then  have  the  contents,  and  two  of  the  dimensions 
given. 

6.  A  circular  cistern  measures  5  ft.  in  diameter,  and  contains 
1000  gallons.     Required,  its  depth?  Ans.  6|4  ft. 

What  is  the  standard  for  Liquid  measure  in  the  United  Slate»t 


342  COMMON    ARITHMETIC.  ScCt.  XVJ 

7.  A  circular  cistern  is  to  be  made  to  contain  1692  gallons,  and 
to  be  8  ft.  deep.     Required,  its  diameter?  Ans.  6  ft. 

Suggestion. — By  putting  some  character,  as  x,  in  the  place  of  the 
dimension  wanted  in  the  last  two  examples,  the  method  of  working 
will  be  at  once  perceived. 

Case  4.     parallelopipeds. 

Obs.  10.  A  Parallelopiped  is 
a  solid  or  prism  having  six  faces,  of 
which  those  opposite  each  other  are 
equal  and  parallel. 


Remark. — A  Cube  is  a  parallelopfped^  each  face  of  which  Is  a  square.  (Sect 
IX,  Art.  1,  Obs.  17.)  A  cominon  brick,  or  chest,  is  a  parallelopiped,  each 
side  of  which  is  a  rectangle. 

To  find  the  surface  of  a  parallelopiped  : 

Obs.  11.  Find  the  surf  ace  of  each  side  separately,  and  add  the 
several  results  together. 

Remark. — The  surface  of  any  solid  bounded  by  plane,  or  straight  edged  sur- 
faces, is  found  by  the  same  Rule. 

1.  Required,  the  surface  of  a  parallelopiped  6  ft.  long,  4  ft.  wide, 
and  4  ft.  deep?  Ans.   128  sq.  ft. 

2.  Required,  the  surface  of  a  solid  4  ft.  9  in.  long,  and  .3  ft.  10 
in.  wide,  and  2  ft.  8  in.  thick?  Ans.  68  sq.  ft.  104  sq.  in. 

To  find  the  solid  contents  of  a  parallelopiped  : 

Obs.  1 2.  Multiply  together  its  length,  breadth^  and  thickness.  (Sect- 
IX.  Art.  2.  Obs.  18.) 

3.  What  is  the  solidity  of  a  parallelopiped  12  ft.  long,  6  ft.  wide, 
and  4  ft.  deep?  Ans.  288  s.  ft. 

4.  Required,  the  contents  of  a  solid  24  ft.  long,  20  ft.  wide,  and 
16  ft.  thick?  Ans.  7680  s.  ft. 

Case  5.     Pyramids. 

Obs.  13.  A  Pyramid  i?  a  solid  which  decreases  gradually  from 
its  base,  until  it  comes  to  a  point  at  the  top.  This  point  is  called  the 
Vertex. 

What  is  a  Parallelopiped?  Give  examples.  How  do  we  find  the  surface  of 
a  Parallelopiped?  What  olh-^r  solidscan  have  their  surfaces  found  by  the  same 
rule?  How  do  we  find  thesolidiiy  uf  a  Parailelo.pi,ed?  Wijat  is  a  Pyramid? 
What  is  the  point  at  the  lop  called? 


Art.  2. 


MENSUBATION    OF    SOLIP?!. 


343 


Pyramids  are  triangular,  quadrangular ,  circular,  <fec.,  according 
as  their  bases  are  triangles,  squares,  circles,  &c.  A  circular  pyra- 
mid is  called  a  Cone. 

Obs.  14.  A  Frdstrum  is  what  remcdvs  cfler  the  top  of  the  pyra- 
mid or  -one  has  heen  taken  off.  The  ;^op  anJ  botjm  of  the  frustrum 
are  called  its  bases. 


Sq.    Pyramid.  Triangular  i'VictiUid. 

Cone. 
If  in  the   above  diagrams,  the  part  0-D  E  F  be  taten  from  the 
triangular  pyramid,  the  remaining  part  ABC-DEF  will  be  the  frus- 
trum. 

Also,  if  the  pari  0-CD  betaken  from  the  cone,  the  remaining 
part  AB-CD  will  be  ihe  frustrum. 

Obs.  15.  The  Altitude  of  a  pyramid,  or  cone,  is  a  line  passing 
through  the  centre  from  the  vertex  to  the  base.  A  line  drawn  from 
the  vertex  perpendicularly  to  one  side,  is  called  its  SlaiU  Heiglith. 
Thus,  OS  is  the  slant  heig  th  of  the  pyramid  0-ABC. 

To  find  the  surface  of  a  pyramid,  or  cone  : 

Obs.   16.      Multiply  the  perimeter  of  its    base  by  half  its    slant 

heighth. 

Demonstration. — The  sides  of  pyramids  are  triangles,  which  the 
slant  heighth  divides  into  right  angled  trian!j;les,  of  which  the  area 
is  found  by  multiplying  he  base  by  half  the  perpendicular.  (Art.  1. 
Obs.  3.)  But  the  sum  of  all  the  ba^es  of  these  triangles  is  the  per- 
imeter of  the  pyramid,  which  multiplied  by  half  the  common  per- 
pendicular, or  slant  heighth  of  the  pyramid  gives  the  surface. 


When  are  pyramiHs  said  to  be  triangular,  qundrangular,  &,c.?     What  is 
Cone?     A  Frustrum?     What  are  the  top  and   bollom  of  the  frustrum  callec 


a 

^„.,. „ ^ called? 

What  is  the  altitude  of  a  Pyramid,  or  Cone?     Its  slant  heighth?     How  do  we 
find  the  surface  of  a  Pyramid,  or  Cone?     Demonstrate  this  rule? 


344  COMMON    ARITHMETIC.  Sect.    XVI 

Again  :  if  the  bases  of  these  triangles  are  bisected  indefinitely, 
they  will  become  a  circle,  and  the  pyramid  will  become  a  cone. 

Hence — The  surface  of  a  cone  is  found  by  the  same  process. 
(Art.  1.  Obs.  9.) 

1.  Required,  the  area  of  the  surface  of  a  triangular  pyramid,  the 
slant  heighth  of  which  is  30  ft  ,  and  each  side  of  its  base  12  ft.? 

Ans.  540  sq.  ft. 

2.  Required,  the  surface  of  a  quadrangular  pyramid,  each  side 
of  which  is  9  ft.,  and  the  slant  heighth  15  ft.?      Ans.  270  sq.  ft. 

3.  Required,  the  area  of  the  surface  of  a  cone  3  ft.  in  diameter, 
at  the  base,  and  its  slant  heigth  12  ft.?  Ans.  56.5488  sq.  ft. 

4.  Required,  the  area  of  the  curved  surface  and  base  of  a  cone, 
the  slant  heighth  of  which  is  12  ft.  11  in.,  and  the  diameter  of  its 
base  4  ft.  7  in.?  Ans.   109  sq.  ft.  70.905  sq.  in. 

To  find  the  surface  of  the  frustrum  of  a  pyramid  or  cone  : 
Obs.  17.    Add  together  the  perimeters  of  the  upper  and  lower  bases, 
and  multiply  half  their  sum  hy  the  slant  heighth. 

Demonstration. — From  the  diagram  under  Obs.  12,  we  perceive 
the  sides  of  the  frustrum  of  a  pyramid  are  trapezoids.  (Art.  1. 
Obs.  5.)  But  the  area  of  these  is  found  by  multiplying  half  the 
sum  of  the  parallel  sides  (or  bases,)  by  their  altitude,  (or  slant 
heighth.)  (Art.  1.  Obs.  6.)  Hence,  as  the  sum  of  the  several  bases 
forms  the  perimeters  of  the  two  bases  of  the  pyramid,  and  their  al- 
titude, or  slant  heighth  is  the  same  in  all,  the  rule  is  evident. 

Again  :  If  the  enc  s,  or  bases  of  these  trapezoids  are  bisected  in- 
definitely, they  will  become  circles,  and  the  frustrum  of  a  pyramid 
will  become  the  frustrum  of  a  cone.  Hence,  the  surface  of  the  frus- 
trum of  a  cone  is  found  by  the  same  process.  (Art.  1.  Obs.  9.) 

5.  Required,  the  convex  surface  of  the  frustrum  of  a  triangular 
pyramid,  of  which  the  upper  base  measures  6  ft.,  and  the  lower 
base  9  ft.  on  a  side,  and  the  slant  heighth  is  12  ft.? 

Ans.  270  sq.  ft. 

6.  Required,  the  convex  surface  of  the  frustrum  of  a  quadrangu- 
lar pyramid,  of  which  the  slant  heighth  is  36  inches,  and  the  upper 
and  lower  bases  14  and  26  inches  on  a  side?  Ans.  20  sq.  ft. 

7.  Required,  the  convex  surface  of  the  frustrum  of  a  cone  of 
which  the  slant  heighth  is  30  ft.,  the  diameter  of  the  upper  base  12 
ft.,  and  that  of  the  lower  base  18  ft.? 

Ans.   1263  sq.ft.   103.68  sq.  in. 

To  find  the  solidity  of  a  pyramid  or  cone  : 

Obs.  1 8.     Multiplg  the  area  of  its  base  by  one-third  of  its  altitude. 

How  do  we  find  the  surface  of  the  frustrum  of  a  Pyramid,  or  cone?  De- 
monstrate this  rule.     How  do  we  find  the  solidity  of  a  pyramid,  or  cone? 


Art.    2.  MENSURATION    OF    SOLIDS.  345 

The  demonstration  of  this  and  the  next  rule  in  this  case,  and  alsO 
of  the  rules  of  the  next  case,  is  too  complex  to  be  understood  with" 
out  a  knowledge  of  Geometry. 

8.  Required,  the  solid  contents  of  a  triangular  pyramid,  whose 
base  is  6  ft.  on  each  side,  and  altitude  25  ft.? 

Ans.   130  s.  ft.  nearly, 

9.  One  of  the  Egyp  ian  pyramids  is  square  at  the  base,  measuring 
720  ft.  on  a  side,  and  its  heighth  is  477  ft.     jfeequired,  i*s  contents? 

Ans.   82425600  s.  ft. 

10.  Required,  the  solid  contents  of  a  cone  20  ft.  in  heighth,  and  5 
ft.  in  diameter  at  the  base? 

Ans.   130.9  s.  ft. 

11.  How  many  times  can  a  conical  vessel  5  inches  in  diameter, 
and  8  inches  deep,  be  filled  from  a  hogshead  of  cider? 

Note. — The  learner  will  recollect  that  a  gallon  of  cider  contains  231  cubic 
inches. 

Ans.  277Vy  times. 

To  find  the  solidity  of  the  frustrum  of  a  pyramid  or  cone  : 

Obs.  19.  Find  the  area  of  each  end,  or  base  ;  Multiply  the  two 
areas  together,  and  extract  the  square  root  of  their  product.  Finally, 
add  together  the  areas  of  the  upper  base,  the  lower  base,  and  the 
root  just  found,  and  multiply  their  sum  by  one-third  of  the  altitude, 

12.  Required,  the  solidity  of  a  block  whose  ends  are  squares, 
each  side  of  the  lower  base  being  4  ft,,  each  side  of  the  upper  base 
3  ft.,  and  its  length  15  ft.? 

Ans.   185  s.  ft. 

13.  Required,  the  solidity  of  a  block  whose  ends  are  triangles, 
the  sides  of  the  top  end  being  2,  2j,  and  3^^  feet,  the  sides  of  the 
lower  end  being  ^\,  4,  and  b\  ft,,  and  its  length  being  12  ft.? 

Ans.  56.5+ s.  ft. 

14.  Required,  the  solid  contents  of  a  glass  in  the  form  of  a  frus- 
trum of  a  cone,  the  diameter  of  the  top  being  4  inches,  and  of  the 
bottom  5  inches,  and  its  depth  9  inches? 

Ans.   143.72823  in. 
,  ^   15.  Required,  the  solid  contents  of  a  mound  in  the  shape  of  a 
frustrum  of  a  cone,  the  diameter  of  the  upper  base  being  15  ft.,  of 
the  lower  base  20  ft.,  and  its  iTeighth  30  ft.? 

Ans.   7264.95  s.    ft. 


How  do  we  find  the  ■olidity  of  the  frustrum  of  a  pyramid  or  cone? 


846 


COMMON    ARITHMETIC. 


Sect  XVI 


Case  6.     Spheres,  or  Globes, 

Obs.  20.  A  Globe,  or  Sphere,  u  a  solid, 
in  which  all  parts  of  the  surf  we  are  equally  dis- 
tant from  a  point  within  called  the  centre ;  as  a 
cannon  ball,  a  ball  of  ^arn,  &c. 


6 

('  1 1  III  I 


^^^j^. 


To  find  the  area  of  the  surface  of  a  Globe,  or  Sphere  : 
Obs.  21.     Multiply  the  circumference  hy  the  diameter  ;  or 
Multiply  the  square  of  the  diameter  5y  3.1416;  or, 
Multiply  the  square  of  the  circumference  hy  .3183. 

1.  Required,  the  area  of  the  surface  of  a  globe   14  in.  in  diam- 
eter? Ans.  4  sq.  ft.  39.7536  sq.  in. 

2.  What  is  the  area  of  the  surface  of  a  sphere  47.124  ft.  in  cir- 
cumference? Ans.  706.86  sq.  ft. 

To  find  the  solidity  of  a  Globe,  or  Sphere  : 

Obs.  22.     Multiply  the  cube,  or  diameter,  hy  .6236  .*  or, 
Multiply  the  surface  hy  one- sixth  of  its  diameter. 

3.  Required,  the  solidity  of  three  globes,  whose  diameters  are 
12,  16,  and  20  inches  respectively? 

Ans  904.':  808;  2144.6656;  and  4188.8  s.  in.  respectively. 

4.  If  the  diameter  of  the  earth  is  7912  miles,  what  is  its  solidity, 
supposing  it  to  be  a  perfect  sphere? 

Ans.  269333411782.8608  s.  miles. 


Case  7.     Tonnage  of  Vessels. 

Obs.  23.  The  first  thing  necessary  in  ascertaining  tonnage,  is  to 
find  the  number  of  cuhic  feet  of  wafer  displaced  by  the  vessel.  This 
is  done  by  finding  the  s  lid  contents  of  the  hull,  which  is  equal  to 
the  product  of  the  length,  width,  smd  depth  of  the  vessel. 

A  body  floating  on  a  fluid  will  displace  as  much  of  the  fluid  as  is 
equal  to  its  own  weight;  and  a  cubic  foot  of  water  weighs  1000  oz. 
Avor.;  therefore,  95  cubic  feet,  (which  is  allowed  for  a  ton  of  2240 
lbs.),  will  weigh  5937^- lbs.;  that  is,  nearly  three  times  the  weight  of 
th    freight  in  water  is  allowed  to  the  ton. 


Wh^ii 


Glul 


How    Jo  wft  fi..d  tht*  surface  of  n   globe,  or 


jspliert-?  How  do  we  find  Its  soiidii}'?  Wliiit.  is  liie  n'st  tiiiitjr  m-CH-syry 
ascertainingtonnajre?  How  is  this  found?  Wliatquntility  of  fluid  willabody 
floating  on  its  surface  displace?  How  much  does  a  cubic  foot  of  water  weigh? 
How  roany  cubic  feat  are  aiiowetl  to  th&ton?     How  muih  will  a  ton  weigh? 


Art.    2.  MENSURATION    OF    SOLIDS.  347 

There  are  two  rules  for  calculating  tonnage — the  OovernmerU 
Rale,  and  the  Carpenter's  Rule. 

Government  Kule  :     For  Di  uble-d^ched   Vessels. 

**  If  the  vessel  be  double-decked,  take  the  length  thereof,  front  the 
fore  part  of  the  main  stem  to  the  after  2)art  of  the  stern  part,  above 
the  zipper  deck ;  the  breadth  thereof  at  the  broadest  pa^t  above  the 
main  toales,  half  of  which  breadth  shall  be  accounted  the  depth  of  such 
vessel;  and  then  deduct  from  the  length  three-ffths  of  the  breadth^ 
multiply  the  remainder  by  the  breadth,  and  ike  product  by  the  depths 
and  divide  this  product  by  95,  the  quotient  whereof  shall  be  deemed 
the  true  contents,  or  tonnage  of  such  ship  or  vessel." 

For  Single-decked   Vessels. 

"  If  the  ship  or  vessel  be  single-decked,  take  the  length  and 
breadth,  as  above  directed,  deduct  from  the  said  length  th.ee-fifths 
of  the  breadth,  and  take  the  depth  from  the  under  side  of  the  deck 
plank,  to  the  ceiling  of  the  hold;  then  multiply  and  divide  as  afore- 
said and  the  qu(»tient  shall  be  deemed  the  tonnage." 

Remark  1. — The  Carpenter's  Ru'e  is  the  same  as  the  Government  Rule,  ex- 
cept that  the  1  of  the  breadth  is  not  deducted  from  the  length,  but  all  is  cal- 
culated. 

2. — The  Keel  of  a  vessel  is  its  main  bottom  in  length  ;  the  beam  is  its  great- 
est width  from  side  to  side  of  the  hull;  and  the  hold,  the  depth  fro...  the  main 
deck  to  t.ie  bollom  of  the  hull. 

2.  Required,  the  tonnage  of  a  single-decked  vessel,  by  both 
rules,  125  ft.  keel,  38  ft.  beam,  and  10  ft.  hold? 

Ans.  By  Gov.  Rule,  208}  T.;  By  Car.  Rule,  500  T. 

Operation.— ^^  of  38  =  22}  ;   125  —  22}  =  102^  ;    ( 102^  X  38 
X  10) -V- 95  =  208}. 
By  Carpenter's  rule  :  (125  X3SX  10)-r-95  =  500. 

2.  Required,  the  tonnage,  by  the  Government  rule,  of  a  vessel 
228  ft.  long,  35  ft.  wide,  and  Hj:  ft.  deep?  Ans.   1086 J  T. 

3.  Required,  the  Government  tonnage  of  a  double-decked  vessel 
300  ft.  long,  and  35  ft.  wide?  Ans.   1798fl  T. 

4.  Required,  the  Government  tonnage  of  a  double-decked  vessel 
379  ft.  in  length,  and  30  ft.  in  width?  Ans.   1710  T. 


Hov/  many  rules  are  there  for  calculating  tonnage?  What  is  the  Govern- 
ment Rule  for  double-dt^cked  vessels?  p'or  single-deck^'d  vessels?  What  is 
the  Carpenter's  rula?  Wnat  is  the  the  keel  of  a  voBsel?  Th»  beamf  Thf 
h»ld» 


'■'  ^' 


848  COMMON    ARITHMETIC.  Sect.    XVI 

Case  8.     Ireegular  Bodies. 

To  find  the  solidity  of  any  irregular  body  wLicli  cannot  be  re- 
duced to  a  regular  form  : 

Obs.  22.  Immerse  it  in  a  vessel  partly  filled  with  water ;  then 
the  solid  contents  of  that  part  of  the  vessel  filled  by  the  rising  of 
the  water,  will  he  the  solidity  of  the  body  immersed. 

This  rule  is  so  evident  that  it  needs  no  demonstration, 

1.  A  stone  of  irregular  form  being  put  into  a  tub  partly  filled 
"with  \vater,  raided  the  water  4^  inches.  Required,  the  solidity  of 
the  stone,  the  diameter  of  the  tub  being  18  inches? 

Ans.   1145.1132s.  in. 

2.  A  lobster  being  put  into  a  bucket  partly  filled  with  water,  raised 
the  water  3  inches.  The  diameter  of  the  bucket  at  the  surface  of 
the  water,  belore  the  lobster  was  put  in,  was  9  inches,  and  the  diam- 
eter at  the  surface  after  the  lobster  was  put  in  was  10  inches.  Re- 
quired, the  solidity  of  the  lobster?  Ans.  212.8434  s.  in. 

3.  Required,  the  solidity  of  a  brush-heap,  which  being  put  into  a 
conical  cistern,  raised  the  water  25  inches,  the  diameters  being  5  ft. 
10  in.  and  6  ft.  3  in.  respectively ?         Ans.  59  s.  ft.  1295|  s.  in. 

Article  3»     Gauging. 

Obs.  1.  The  metJiod  of  finding  the  contents  of  any  vessel  in  gal- 
lons, bushels,  dx.,  is  called  Gauging. 

To  find  the  number  of  bushels  which  any  vessel  will  contain  : 

Obs.  2.  I'ind  the  solid  contei.'ts  of  the  vessel  in  feet ;  then  multiply 
this  by  45,  and  divide  the  product  bg  56'. 

Demonstration. — In  a  solid  foot  are  1728  solid  inches  ;  (see  table 
Cubic  Measure;]  in  a  bushel  tin  re  are  2150.4  solid  inches  ;  (see  ta- 
ble Dry  Measure;)  therefore  a  bwshel  contains  \\%\\~z=i  \\  solid 
feet.  Hence,  the  vessel  will  contain  as  many  bushels  as  U  is  con- 
tained in  its  solid  contents.     That  is,  we  divide  by  | J. 

a.  For  heaped  measure  use  9  and  14  instead  of  46  and  h^^  be- 
cause  1411  =  V. 

Ex.  1.  How  many  bushels  will  a  box  hold  that  is  6  feet  long,  4 
ft.  wide,  and  6  ft.  deep?  Ans.   II54. 

2.  How  many  bushels  will  a  gum  contain  which  is  12  ft.  deep, 
and  4  ft.  in  diameter?  Ans.   121,176. 

How  do  we  find  the  solidity  of  irregular  bodies?  Wliat  is  gruaging?  How 
do  we  find  the  number  of  bushels  any  vessel  will  contain?  How  do  we  pro- 
ceed with  heaped  measure?     Why? 


Art.    3.  MENSURATION    OF    SOLIDS.  349 

3.  How  many  bushe's  would  a  gum  contain,  which  is  8  ft.  deep, 
and  measures  6  ft.  in  diameter  at  the  top,  and  5  ft.  in  diameter  at 
the  bottom?  Ans.   153.155. 

4.  How  many  bushels  of  apples  can  be  put  in  a  bin  35  ft.  long, 

6  ft.  wide,  and  5  ft.  deep?  Ans.  e'/S. 

Note. — The  learner  will  observe  that  this  is  heaped  measure. 

5.  How  many  bushels  of  potatoes  can  be  put  in  a  bin  25  ft.  long, 

7  ft.  wide,  and  6  ft.  deep?  Ans,  675. 

6.  A  bin  is  6  ft.  wide,  and  6  ft.  deep ;  how  long  must  it  be  to 
contain  810  bushels  of  apples?     To  contain  810  bushels  of  wheat? 

Ans.  35  ft.  for  apples;  and  28  ft,  for  wheat. 

Remark. — When  it  is  desired  to  obtain  the  contents  in  barrels,  we  divide 
the  r'-sult  obtained  by  5.  If  corn  is  the  thing  measured,  and  it  is  in  the  ear, 
we  divide  by  10  ;  if  in  the  husk,  by  20. 

7.  How  many  gallons  of  water  will  a  conical  cistern  contain,  that 
is  7  ft.  9  in.  deep,  and  the  diameter  5  ft.  6  in.  at  the  bottom,  and  4 
ft.  9  in.  at  the  top?  Ans.   1217.0636. 

Note. — We  find  the  solid  contents  in  inches,  and  divide  this  by  231,  because 
231  cubic  inches  make  a  gallon.  (See  table  Wine  measure.) 

To  find  the  numb  ir  of  wine  and  beer  gallons  contained  in  a  cask, 
or  barrel : 

Obs.  3,  Take  the  dimensions  of  the  cask  in  inches  :  that  is,  the 
diameter  at  the  bung  and  head,  and  the  length  of  the  cask. 

If  the  staves  are  much  curved,  multiply  the  difference  between  the 
bung  and  head  diameter  by  .7;  if  but  little  curved,  hy  .6;  if  they 
are  of  a  medium  curve,  by  .65  ;  or.  if  they  are  almost  straight 
by  .55  ;  in  either  case,  add  the  product  to  the  head  diameter,  and 
the  sum  wUl  be  the  mean  diameter,  and  the  cask  zvill  be  reduced  to 
a  cylinder. 

Multiply  the  square  of  the  mean  diameter  by  the  length  of  the 
cask,  and  thai  product  by  34  for  wine  measure,  and  28  for  beer 
ineasure,  and  point  off  four  decimals  in  the  p)roduct  ;  the  result  will 
be  in  gallons,  and  decimals  of  a  gallon. 

Demonstration, — The  demonstration  of  this  rule,  until  the  cask  is 
reduced  to  a  cylinder,  cannot  be  understood,  without  a  knowledge 
of  the  higher  departments  of  mathematics.  After  it  is  reduced  to  a 
cylinder,  we  find  its  solidity  by  multiplying  the  square  of  the  mean 
diameter  by  .7854,  and  its  length;  (Ait,  1.   Obs.  11,  and  Art.  2. 

How  do  we  find  the  contents  in  barrels?  How  do  we  find  the  number  of 
wine  and  beer  gallons  contained  in  a  cask  or  barrel?     Demonstrate  this  rule. 


850  COMMON      f^RlTHMETIO.  Scct.    XVII 

Obs.  7.)  and  this  is  reduced  to  gallons  by  dividing  by  231  for  wine 
measure,  and  282  for  beer  measure.  (Obs.  3.)  Thus,  in  wine  mea- 
sure, ,  7854  is  a  multiplier,  and  231  is  a  divisor  ;  that  is,  we  multi- 
ply by  'Vg-V  ;  but  '%%\^  =  .0J34.  Also,  in  beer  measure,  we 
multiply  by  *VjV,  and  'Vg-V  ==■  -^028  nearly.  Hence,  this  part 
of  the  rule  is  evident. 

Remark. — Guaging  of  casks  by  computation  is  of  but  little  ufility,  as  it 
[6  now  mostly  done  by  a  rod,  with  calculations  already  made,  in  tabular  form. 

8.  Required,  the  contents,  in  wine  and  beer  gallons,  of  a  cask, 
of  which  the  staves  arc  much  curved,  the  diameter  at  the  head  be- 
ing 20  inches,  and  at  the  bung  30  inches,  and  its  length  40  inches? 

Ans.  99.144  wine  gallons  ;  81.648  beer  gallons. 

9.  How  many  wine  and  beer  gallons  in  a  cask  45  inches  in  length, 
the  bung  diameter  30  inches,  and  the  head  diameter  25  inches,  the 
staves  being  a  medium  curve? 

Ans.  122.1035-f-  wine  gallons  ;   ir.0.5558  beer  gallons. 

10.  Required,  the  number  of  wine  and  beer  gallons  in  a  cask  of 
which  the  staves  are  nearly  straight,  and  the  dimensions  as  follows  : 
bung  diameter,  20  inches;  head  diameter,  16  inches;  length,  50 
inches'?  Ans.  56.3100  wine  gallons  ;  46.3736  beer  gallons. 

11.  How  many  wine  and  beer  gallons  will  a  hollow  cylinder  con- 
tain, which  is  4  feet  deep,  and  1  ft.  6  in.  in  diameter? 

Ans.  52,8768  wine  gallons  ;  43.6456  beer  gallons. 


SECTION  XVH. 

PHILOSOPHICAL   AKD  MISCELLANEOUS  QUESTIONS. 

Article   1.     The  Mechanical  Powers. 

Obs.  1.  The  Mechanical  Powers  consist  of  six  simple  insti'w 
me)its,viz':  the  Lever,  the  Wheel  and  Axle,  the  Pulley,  the  Inclined 
Plane,  the  Wedge,  and  the  Screw. 

Obs.  2  A  Machine  is  an  instrument  cornhining  one  or  more  of 
the  mechanical  poioers.  It  may  be  very  simple,  as  a  pin ;  or  very 
complicated,  as  a  steam-engine. 

Obs.  3.  The  Power  is  the  body  which  applies  the  force  ;  ar.d 
the  Weight  is  the  body  which  resists  the  force. 

What  are  th«  Mechanical  Powera?  What  is  a  Machinef  The  power? 
Tbe  weight? 


Art.     1.  THE    MECHANICAL    POVVKRS.  51 

Case  1.     The  Leveh. 

Obs.  4.  The  Lever  is  simpbj  a  bar,  used  in  raising  weights. 
It  is  moved  about  a  fixed  point,  called  its  Fulcrum,  or  Prop. 

■  Remark. -There  are  three  kinds  of  levers.  Force  exerted  by  the  first  kind  is 
called  prijtna-.  In  this  the  weight  and  power  are  at  the  ends  of  the  lever, 
*nd  the  fulcrum  between  them.  Force  exerted  by  the  second  kind  is 
called  lifting.  In  this  the  power  and  fulcrum  are  at  the  ends  of  the  lerer, 
and  the  weight  between  them.  Force  exerted  by  the  third  power  is  called 
lifting,  or  raising.  In  this  the  weight  and  fulcrum  are  at  the  ends  of  the 
lever,  and  the  power  between  them.  Raiding  a  ladder  to  the  roof  of  a 
house  is  an  example  of  this  kind  of  lever. 

2. — The  arms  of  the  lever  are  the  parts  between  the  fulcrum  and  its 
ends. 

Obs.  5.  A  power  and  weight  acting  upon  the  arms  of  a  lever 
will  balance  each  other,  when  the  product  arisiiig  from  multiplying 
the  power  by  Us  distance  from  the  fulcrum,  is  equal  to  the  product  ari- 
sing from  multiplying  the  weight  by  its  distance  from  the  fulcrum. 

By  keeping  this  principle  in  view,  we  can  always  find  any  one  of 
these  four  things  :  the  power,  the  weight,  the  distance  of  the  weight 
from  the  fulcrum,  and  the  distance  of  the  power  from  the  fulcrum, 
if  the  other  three  are  given. 

Note — By  this  rule,  the  power  and  weight  exactly  balance;  in  order  to 
raise  the  weight,  there  must  be  a  small  increase  of  power.  It  is  a!?o  cus- 
tomary in  practice,  when  estimating  the  mechanical  powers,  to  deduct  one' 
third  (or  friction. 

Ex.  1.  Suppose  a  man  weighing  140  lbs.  rests  on  a  lever  8  feet 
long,  what  weight  will  he  balance  on  the  other  end,  supposing  the 
fulcrum  to  be  6  inches  from  the  weight?  Ans.  2100  lbs. 

2.  Given,  the  weight,  d  stance  of  the  weight  from  the  fulcrum, 
and  the  distance  of  the  power  from  the  fulcrum,  in  the  last  example, 
to  find  the  power?  Ans.   140  lbs. 

3.  Suppose  a  power  of  150  lbs.  be  applied  to  a  lever,  16  ft.  fr^ra 
the  fulcrum,  what  weight  will  it  balance  at  the  other  end  of  the 
lever,  which  is  4  inches  from  the  fulcrum?  Ans.  6750  lbs. 


What  is  a  lever?  How  many  kinds  of  levers  are  there?  What  is  force  ex- 
erted by  the  first  kind  called?  How  are  the  power,  weight  nud  fnlcrvmi  ar- 
ranged with  respect  to  each  other?  What  is  force  exertea  by  the  secouii  kind 
of  lever  culled?  How  are  the  power,  weight,  and  fulcrum  arranged  with  res- 
ppct  to  each  otiier?  What  is  force  exerted  by  the  third  kind  of  lever  called? 
How  are  the  weight,  power,  and  fulcrum  arrnuged  with  respect  to  each  other? 
Give  an  example  of  this  kind  of  lever.  What  are  the  arrns  of  a  lever?  When 
will  the  power  and  weight  acting  upon  the  arms  of  a  lever  balance  each  other? 
What  is  necessary  iu  order  to  raise  the  weight?  What  deduction  is  made  for 
friction  In  the  praotisal- application  of  the  m^ehantcul  powers? 


352  COMiMON    ARITHMETIC.  Scct.    XVII 

4.  By  a  lever  30  ft.  long,  what  weight  will  400  lbs.  sustain,  the 
weight  being  6  inches  from  the  fulcrum?  Ans.  23600  lbs. 

'    5.   By  a  lever  12  ft,  long,  what  weight  will  200  lbs.  balance,  the 
power  being  applied  18  inches  from  the  fulcrum?     Ans.  284  ^^^• 

Case  2.     The  Wheel  and  Axle, 

Obs,  6.  The  Wheel  and  Axle  consists  of  two  wheels,  one  of 
which  is  larger  than  the  other,  hut  the  smaller  one  passes  through 
the  larger,  and  thus  both  have  a  common  centre  on  which  they  turn. 

The  power  is  applied  to  the  circumference  of  the  larger  wheel,  and 
the  weight  tj  that  of  the  smaller  wheel  (or  axle),  by  means  of  cords. 

Obs.  7.  The  power,  when  applied  to  this  machine,  will  exactly 
balance  the  weight,  when  the  product  arising  from  multip)lying  the 
power  by  the  diameter  of  the  wheel,  is  equal  to  the  product  of  the 
weight  by  the  diameter  of  the  axle. 

Hence,  if  any  three  of  these  are  given,  the  other  can  be  easily 
found. 

1.  If  the  diameter  of  the  wheel  is  4  ft.,  and  the  diameter  of  the 
axle  3  inches,  what  weight  on  the  axle  will  balance  125  lbs,  ap- 
plied to  the  wheel?  Ans.  2000  lbs, 

2.  If  the  diameter  of  the  wheel  is  6  ft.  8  in.,  and  the  diameter  of 
the  axle  10  inches,  what  power  must  be  applied  to  the  wheel  to  bal- 
ance 2600  lbs.  at  the  axle?  Ans.  325  lbs. 

3.  If  the  diameter  of  the  wheel  is  5  ft.,  the  power  224  lbs.,  and 
the  weight  2688  lbs.,  what  is  the  diameter  of  the  axle? 

Ans,  5  inches. 

4.  If  the  diameter  of  the  axle  is  8  inches,  the  power  340  lbs., 
and  the  weight  2975  lbs.,  what  is  the  diameter  of  the  wheel? 

Ans.  5  ft.  10  in. 

5.  Suppose  there  are  two  wheels  :  one  6  ft.  9  in.  in  diameter, 
with  an  axle  9  inches  in  diameter,  and  the  other  5  ft.  5  in.  in  diam- 
eter, with  an  axle  6  inches  in  diameter ;  if  the  power  cord  of  the 
smaller  be  attached  to  the  axle  of  the  larger,  what  weight  at  the 
axle  of  the  smaller  wheel  can  be  supported  by  275  lbs.  at  the  power 
cord  of  the  larger?  Ans.   26812^  lbs. 

Case  3.     The  Pulley, 

Obs.  8.  A  Pulley  is  a  grooved  ivheel,  which  turns  about  its 
axis  by  means  of  a  cord  passing  over  it.  It  may  be  either  simple 
or  compound,  fixed  or  movable. 

What  is  a  Wheel  and  Axle?  How  is  the  power  applied?  The  weight? 
When  will  the  weight  and  power  exactly  balance?     \Vhat  is  a  Pulley? 


Art.    1.  THE    MECHANICAL    POWERS.  353 

Obs.  9.  The  simple  pvlley  consists  of  a  single  wheel  and  its  cord, 
to  one  end  of  which  the  weight  is  attached,  and  to  the  other  end  the 
power.  No  advantage  is  gained  from  this  machine,  as  the  power 
and  weight  must  always  be  equal.  When  a  number  of  pulleys  are 
used,  one  half  of  them  are  mova  le,  and  the  whole  is  called  a  sys- 
tem of  pulleys.  It  is  evident  that  the  weight  is  divided  between  the 
number  of  strings,  as  there  are  always  double  the  number  of  mov- 
able pulleys  :  Hence — 

Obs.  10.  The  weight  divided  hy  twice  the  number  oj  movable 
pullieSy  will  give  the  power ;  and  the  power  multiplied  by  twice  the 
number  of  movable  pulleys,  will  give  the  weight. 

1.  What  weight  will  be  balanced  by  a  power  of  200  lbs.  attached 
to  a  cord,  which  passes  over  4  movable  pulleys? 

Ans.   1600  lbs, 

2.  What  power  must  be  applied  to  a  cord  passing  over  6  movable 
pullies  to  balance  3600  lbs.?  Ans.  300  lbs. 

3.  What  number  of  movable  pulleys  would  be  necessary  to  raise 
a  weight  of  1000  lbs,  by  a  power  of  125  lbs.?  Ans.  4. 

4.  If  a  cord  that  passes  over  6  movable  pullies  be  attached  to  an 
axle  4  inches  in  diameter,  and  if  the  wheel  is  6  feet  in  diameter, 
what  weight  can  be  raised  by  the  pulley,  by  applying  200  lbs.  to 
the  wheel?  Ans.  36000  lbs. 

Case  4.     The  Inclined  Plane. 

Obs.  11.  An  Incunkd  Plane  «s  a  plane  surface  tvhich  inclines 
to  or  fr^m  the  earth.  Thus,  a  board  with  one  end  on  the  ground, 
and  the  other  end  on  a  block  of  wood,  forms  an  inclined  plane. 

Obs.  12.  The  power  when  applied  to  this  machine  will  exactly 
balance  the  weight,  when  the  weight  multiplied  by  the  heighth  of 
the  plane,  is  equal  to  the  power  multiplied  by  its  length. 

1.  If  an  inclined  plane  is  150  ft.  long,  and  15  ft.  high,  what 
weight  will  200  lbs.  sustain?  Ans.  2000  lbs. 

2.  If  an  inclined  plane  is  80  ft.  long,  and  12  ft.  high,  what  power 
will  sustain  2880  Jbs.?  Ans.  432  lbs. 

3.  What  power  would  be  required  to  draw  a  train  of  cars  weiglf* 
ing  50000  lbs.  up  ati  inclined  plane  3  miles  in  length,  and  100  ft.m 
heighth?  Ans.  3 15|- lbs.  nearly. 

What  is  a  Simple  Pulley?  What  is  a  system  of  pulleys?  How  is  the  weight 
divided?  To  what  are  these  equal?  How  do  we  find  the  power?  The  weight? 
What  is  an  inclined  plane?  Give  an  example.  When  will  the  power  balance 
the  weight  in  this  machine? 


354  COMMON    ARITHMETIC.  ScCt.    XVII 

Case  5.     The  Wedgf. 

Obs.  13,  The  Wedge  may  he  regarded  as  two  ivclined  planes 
placed  base  lo  base. 

Obs.  14.  From  Obs.  12  we  conclude  tbe  force  will  balance  the 
power  when  ike  product  a  ising  from  multiplying  the  force  produced  at 
the  side,  by  the  breadth  of  the  head  of  the  wedge,  is  equal  to  that 
of  multiplying  the  power  acting  against  the  head  by  the  length  of 
its  side. 

Note  1. — This  force  only  respects  one  side  of  the  wedge  ;  when  the  force 
against  both  sides  are  required,  only  half  the  breadth  of  the  head  must  be 
taken  into  consideration. 

2, — Tn  applying  the  wedge,  the  friction  is  at  least  eqnal  to  the  force  to  be 
overcome  ;  and  therefore  not  less  than  one-half  of  the  power  is  lost,  for  which 
no  allowance  is  made  in  the  above  rule.  The  advautage  ef  the  wedge  arises 
from  the  force  of  percussion  or  blow  with  which  it  is  struck,  which  is  much 
greater  than  that  of  any  dead  weight  or  pressure,  such  as  is  usually  employed 
on  the  other  mechanical  powers. 

1.  What  weight  of  force  would  be  effected  on  either  side  of  a 
wedge,  the  head  of  which  is  3  inches  broad,  and  the  side  15  inches 
long,  by  apower  of  85  pounds?  Ans.  425  lbs 

2.  What  power  must  be  applied  to  a  wedge  ■  8  inches  long  and 
6  inches  broad  at  the  head,  to  effect  a  force  of  1575  lbs.,  allowing 
no  loss  for  friction?  If  the  friction  is  equal  to  the  force  to  be 
overcome,  what  power  must  be  employed? 

.  {    525  lbs.  allowing  no  loss  for  friction. 

1050  lbs.  allowing  loss  for  friction. 

Case  6.     The  Screw. 

Obs.  15.  The  Screw  may  be  considered  as  a  thread,  or  groove, 
running  spirally  around  a  cylinder. 

Remark. — The  principle  of  the  screw  is  the  same  as  that  of  the  inclined 
plane;  the  distance  between  the  Ihreads  being  the  hei^hih,  and  the  circiimfer- 
eece  of  the  cylinder  the  base  of  the  plane. 

Obs,  16,  The  power  when  applied  to  tl- is  machine  will  exactly 
balance  the  weight,  vjhen  the  power  multiplied  by  the  circumference 
of  the  circle  described  by  the  power,  is  equal  to  the  uc'ioht  multiplied 
by  the  distance  between  the  threads. 

What  is  a  wedge?  Wlien  will  the  force  l>;t!rt!!ce  Ih.e  power?  What  has  this 
force  reference  to?  When  the  force  against,  lioth  sides  is  required,  how  do  we 
proceed?  What  part  of  the  power  is  lost  in  appUing  the  wedge  fn  p'Hctical 
purposes?  How  so?  What  advnntage  then  I'.as  the  wedjr'^?  What  is  a 
screw?  What  is  the  principle  of  the  screw?  Show  the  comparison.  When 
, will  the  power  balance  the  weight  when  ap[)!ied  to  this  machine? 


Art.     1  .  THE    MECHANICAL    POWERS.  355 

Hence,  if  any  of  these  four  quantities  are  given,  the  other  can 
be  easily  found. 

1 .  What  weight  may  be  supported  by  a  screw,  the  lever  of  which 
is  6  feet  long,  and  the  distance  between  the  threads  of  a  screw  \  of 
an  inch,  by  a  power  of  125  lbs.?  Ans.   1 13097. 6  lbs. . 

The  learner  will  observe  that  the  length  of  the  lever  is  but  half  the 
diameter  of  the  circle. 

2.  What  power  would  be  required  to  balance  12000  lbs.,  if  the  dis- 
tance between  the  threads  of  the  screw  was  |-  of  an  inch,  and  the 
length  of  the  lever  8  feett  Ans.   12.43  lbs: 

Case  7.     Machinery. 

Obs,  17.  Bi/  Machineky  is  understood  ike  connection  of  two  or 
more  of  the  mechanical  powers,  by  means  of  belts,  bands,  cogs,  d'c. 

Remark. — The  velocities  of  wheels  are  to  each  other  as  their  like  dimensions.  ^ 

1.  A  cog  wheel  150  inches  in  diameter,  runs  into  another  but  12 
inches  in  diameter.  Required,  the  number  of  revolutions  the  small- 
er wheel  makes  per  minute,  the  larger  performing  8?     Ans.   100. 

Operation.— 12  :  150  :  :  8  :  100.   ' 

Or,  whilst  the  larger  wheel  makes  1  revolution,  the  smaller  one 
makes  150  -M2  =  12^ ;  and  12^  X  8  =  100,  as  before. 

2.  A  wheel  56  inches  in  diameter  is  connected  with  another  6     , 
inches  in  diameter,   and  this  is   attached  to  a  shaft,  on  the  end  of 
which  a  wheel  42  inches   in  diameter  is  joined  to  another  4  inch  s 

in  diameter.     Required,  the  velocity  of  the  latter  wheel  per  minute 
the  larger  one  making  50  revolutions.  Ans.  4900  revolutions. 

3.  How  many  revolutions  will  a  spindle  2  inches  in  diameter 
make  per  minute,  which  is  connected  with  a  drum  4  ft.  in  diameter, 
that  performs  40  revolutions  per  minute?  Ans.  960. 

4.  A  belt  connects  a  drum  5  feet  in  diameter,  making  50  revolu- 
tions per  minute,  with  a  spindle  3  inches  in  diameter.  What  is  the 
velocity  of  the  spindle  per  rauiute?  Ans,    1000  revolutions. 

5.  If  a  drum  5  feet  in  diameter  performs  40  revolutions  per  min- 
ute, what  is  the  diameter  of  a  cylinder  connected  with  it  that  per- 
forms 600  revolutions  per  minute?  Ans.  4  inches. 

6.  If  the  spindle  of  a  com  non  spinning  wheel  is  \  of  an  inch  in 
diameter,  and  the  rim  is  4  ft,  9  in.  in  diameter,  how  many  revolu- 
tions will  the  spindle  perform  whilst  the  rim  performs  15? 

Ans.  3420. 

What  is  understood  by  Machinery?  What  is  the  ratio  of  the  velocities  of 
wheels  to  each  other? 


356 


COMMON    ARITHMETIO. 


Sect.  XVII 


Article  2.     Method   of  keeping   Books,   Forms  of  Notes,   &c, 

Remakk. — It  is  not  our  design  to  present  here  an  elaborate  treatise  on 
book-keapinjf.  We  shall  only  give  a  form  or  two  for  the  benefit  of  those 
whose  entries  are  few,  and  wno  iiave  but  little  business  to  transact.  Those 
who  desire  further  information  are  referred  to  treatises  on  book-keeping, 
where  they  will  find  the  subject  treated  upon  more  fully. 

Obs.  1.  It  is  necessary  that  every  person  should  have  some  reg- 
ular, systematic  method  of  keeping  his  accounts,  because  the  law 
requires  in  all  cases  of  dispute,  that  the  book  in  which  the  charges 
were  originally  made,  be  produced  before  the  court,  and  legal  evi- 
dence given  of  their  correctness. 

The  following  form  is  recommended  for  farmers  and  mechanics, 
whose  entries  are  but  few,  and  generally  made  at  the  close  of  the 
day  or  week.  It  consists  in  writing  the  name  of  the  person  on  the 
left  hand  page.  Dr.  (debtor)  for  the  sum  he  is  owing  you,  and  on 
the  right  hand  page,  Cr.  (creditor)  by  the  sums  you  owe  him.  The 
difference  between  the  Dr.  and  Cr.  sides  will  at  once  show  how 
much  is  due  from  one  to  the  other. 

Henry  Worthy,  Dr.      Henry  Worthy,  Cr. 


l«4d 
Jan.  1. 
"     4 
«     7 

"     8 


For  8  bushels  of  Wheat, 

at  $1  per  bushel,. . . . 
For    one    day's   work, 

Chopping,  .'. 

For  12    lbs.  butter,   at 

IB^cts.  perlb 

For  2  doz.  Eggs,  at  I2i 

cts.  per  doz 


* 

cts. 

1«4« 
Jan.  3. 

8 

00 

75 

«     6 
«    7 

2 

00 
25 

"    8 

in 

00 

1 

By  Cash 

By  Goods 

By  Cash 

By  balance,  being  the 
amount  now  due  me. 


CIS. 

00 
75 
00 

25 
00 


184S 
Jan.  9. 


For  bal'nce  due  me  from 
old  account 


25 


Note. — It  is  immaterial  whether  we  place  the  debits  and  credits  both  on  the 
same  page  as  above,  or  on  opposite  pages.  The  same  rules  are  observed  in 
both  cases  ;  that  is,  write  the  name  of  the  person  on  the  left  hand  page,  Dr., 
and  on  the  right  hand  page,  Cr.     The  pages  should  be  ruled  as  above. 

Obs.  6.    The  following  remarks  may  also  be  of  use  to  the  learner: 

a.  The  person  who  receives  any  thing  from  you  is  Dr.  to  you 
for  the  amount  he  receives  from  you;  and  the  person  from  whom 
you  receive  any  property  is  Cr.  by  the  amount  you  receive  from 
him. 

b.  Places  of  residence  should  be  named  when  they  are  not  the 
same  as  that  where  the  book  is  kept.  Also,  if  different  persons  bear 
the  same  name,  it  is  best  to  designate  the  occupation  or  particular 


Art.  2. 


BOOK-KEEPING,    &C. 


357 


place  of  residence.     The  name  of  the  person  who  owns  the  book, 
and  his  place  of  residence  should  be  written  on  the  first  page. 

c.  The  date  of  each  transaction  should  be  written  against  the 
entry  of  it  in  the  column  for  that  purpose. 

d.  When  the  account  of  any  person  is  closed  To  (or  For)  Bal- 
ance, you  are  debtor  to  him,  and  when  it  is  closed  By  Balance,  he  is 
debtor  to  you  for  the  amount  of  balance  ;  and  in  opening  a  new  ac- 
count, he  must  be  credited  By  Balance,  or  made  Dr.  To  (or  For,  Bal- 
ance from  one  account  to  another,  the  balance  always  crossing  from 
the  Dr.  to  the  Cr.  side,  or  from  the  Cr.  to  the  Dr.  side,  in  passing 
from  the  old  account  to  the  new. 

e.  "When  a  person  has  dealings  with  several  individuals,  he  should 
have  an  index  to  his  book,  in  which  the  name  of  every  person  with 
whom  he  transacts  business  should  be  written  under  its  nitial,  and 
the  page  where  the  account  is  kept  should  be  noted  down. 

/,  Care  should  be  taken  to  preserve  a  book  free  whom  blots,  and 
mistakes,  and  every  entry  should  be  made  in  a  full,  bold  and  legi- 
ble hand. 

Remarks. — For  drovers,  and  farmers  who  raise  considerable  stock,  or  grain, 
and  are  in  the  habit  of  receiving  and  paying  out  money,  frequently,  the  follow- 
ing form  is  recommended: 


Cash  Account, 


Dr.        Cash  Account, 


Cr. 


jI7 

(( 

4' 

« 

9 

« 

12 

(( 

30 

To  Cash  on  hand  . . . . 
"    Sales  of  Cattle  . . 
"    Sales  of  Wheat  .. 
*«  Cash  recjd  of  Tho- 
mas True   on   his 

note 

"    Cash  rec'd  for  hor- 
ses   


cts.     Ie4» 


200    00 
75    50 


50 


15 


00 


75 


455    75 


767    00 


for 


Jan. '7  By  Cash     paid 

Goods 

Cash  paid  for  work 
Cash  p'd  for  Cattle 
Cash    paid    for 

wheat 

Balance,  being  the 
cash  on  hand. . . 


u  1 
«  25 

•*  28 

«  30 


$    cts. 


300 
15 
66 


00 
25 
75 


20]  00 
365J  00 
767  i  00 


1848 


Feb.  1    To  bal.  from  oH  acc't, 


356    00 


Obs.  3.     The  learner  will  now  notice  the  following   suggestions* 


a.  The   Cash-Book  is  kept  by  making  Cash  Dr.  to  what  is  on 
hand,  and  what  is  received,  and  Cr.  by  whatever  is  paid  out. 

b.  The  excess  of  the  Dr.  side  over  the  Cr.  side  ought  always  be 
€qual  to  the  cash  on  hand. 

c.  When  the  cash  on  hand  is  counted,  it  should  be  entered  on 


358  COMMON    ARITHMETIC.  Scct.    XVII 

the  Ct.  side;  but  when  a  balance  is  struck,  the  cash  on  hand  should 
be  entered  on  the  Dr.  side. 

Note. — Ornament  and  perspicuity  being  the  object  of  many  accountants,  they 
often  use  cap'Ual  Letters,  noiw'whsVAivWng  the  rules  ihey  Irequently  see  to  the 
coutrary.  Tney  likewise,  seldom  use  the  preposition  of,  wriiing  12  lbs.  Butter, 
6  yds.  Sheeting,  &c.,  instead  of  12  lbs.  of  Butter,  6  yds.  of  Sheeting,  &:c. — 
But  tlie  learner  need  observe  but  one  particular  viz:  to  have  a  good  pen, 
and  clean  fingers,  and  to  preserve  his  book  from  blots,  mistakes,  "and  every 
thing  else  but  his  entries. 

FORMS  OF  ORDERS,  RECEIPTS,  AND  NOTES. 

ORDERS. 

Columbus,  June  7th,  1848, 

Mr.  Silas  Broivn — Please  pay  the  bearer,  Mr.  Samuel  Jones^  six 
dolhirs  and  twenty  cents,  and  charge  the  same  to  my  account. 

Thomas  Henry. 

Newark,  Sept.  1st,   1848. 

Mr.  Simon  Dialer — Please  pay  Mr.  Chares  Lane  such  goods  as 
he  may  call  for,  not  excee  ling  the  sum  of  fifteen  dollars,  and  charge 
the  same  to  your  h.tmble  servant.  Thomas  Thrifty. 

Remark  1. — If  the  persons  on  whom  the  order  is  made  resides  in  another 
town  from  the  one  in  which  the  order  is  dated,  the  name  of  the  town  in  which 
they  reside  should  be  written  with  their  names. 

RECEIPTS. 

Genoa,  Oct.  3d,  1848. 

Received  of  Henry  Martin  two  dollars  in  full  of  all  accounts. 

Thomas  Stone. 

Received  of  Moses  Hale  five  dollars  in  full  of  all  demands. 
Dublin,  Dec.  4th,  1848.  Peter  Goodrich, 

Remark  2. — When  a  receipt  is  given  "  in  full  of  all  accounts,"  it  cuts  off 
only  th^*  claims  to  accounts  ;  but  when  it  is  given  "  in  full  of  all  demands,"  it 
cuts  off  all  cldims  of  any  kind. 

Received  of  Thomas  Mosely,  five  dollars  and  fifty  cents  on  his 
note  for  ten  dollars,  dated  Sept.  2nd,  1847. 

Charles  Wakefield. 

Chillicothe,  June  6th,   1848. 
Received  of  WiUiGm  Jones  twelve  dollars,  to  pay  on  the  account 
of  John  Smithi  Henry  Markley. 


Art.    2.  BOOK-KEEPING,   &c.  359 

DUE-BILLS. 

Obs,  4.  A  Due-bill  is  a  mere  pledge  to  pay  a  certain  amount 
of  money  or  other  property  therein  specified,  in  consideration  of  an 
equivalent  specified. 

Form  of  a  Bill. 
Due  Isaac  Dealer,  or  bearer,  twelve  dollars  for  value  received. 
Columbus,  Dec.  4th,   1847.  Jacob  Faithful. 

Note. — If  payment  is  to  be  made  in  any  thing  besidea  money,  it  should 
be  so  specified  in  tlie  due-bill. 

*     '  NOTES. 

No.  1.     Payable  to    Order. 
$200.  Columbus,  Feb.  2nd,   1848. 

For  value  received,  on  demand,  I  promise  to  pay  Abraham  Drive- 
all,  or  order,  two  hundred  dollars. 

James  Constance. 

No.  2.     Note  payable  to  Bearer, 

8500.  Columbus,  March  4th,  1848. 

For  value  received,  on  or  before  the  first  day  of  February  next,  I 
promise  to  pay  Joseph  Goodwill,  or  bearer,  five  hundred  dollars, 
with  interest  from  date. 

Ezra  Fairface. 

No.  3.     Note  by  two  Persons. 
$75.  Cleveland,  April  3d,  1847. 

One  year  after  date,  for  value  received,  we,  jointly  and  severally 
promise  to  pay  Charles  Good,  or  bearer,  seventy-five  dollars,  with 
interest  after  thirty  days. 

Peter  Trueman. 

Attest :  Henry  Jones,  Samuel  Albright. 

REMARKS    ON   NOTES. 

1.  The  drawer  or  maker  of  the  note,  is  the  one  who  signs  it ;  the 
holder  of  the  note  is  the  one  who  has  possession  of  it. 

2.  No  note  is  negotiable  or  transferable,  unless  the  words  *'or 
order,'*  or  the  words  "or  bearer'*  are  inserted  in  it. 

3.  When  the  holder  o!  a  negotiable  note  payable  to  order, 
(Note  1 )  wishes  to  transfer  it,  he  must  endorse  it ;  that  is  write  his 
name  on  the  back  of  it.     The  holder  of  the  note  is  then  authorized 


360  COMMON    ARITHMETIC.  Scct.    XVII 

to  collect  it  of  the  drawer^  but  if  the   drawer   refuses  to  pay  it,  he 
can  collect  it  of  the  endorser. 

4.  When  a  note  is  given  with  surety,  the  surety  is  not  re- 
sponsible only  through  the  inability  of  the  principal,  or  drawer. 

5.  When  a  note  is  made  payable  to  bearer,  (Note  2.)  any  person 
who  has  the  note  can  collect  it  of  the  drawer. 

6.  Th«  words  '-J^or  value  received"  should  be  written  in  every 
note,  and  the  amount  of  the  note  should  always  be  written  in 
words. 

7.  Every  note  should  be  made  payable  on  demand,  or  at  some 
specified  time. 

8.  When  the  time  expires  for  which  the  note,  was  given,  the 
note  will  draw  interest,  although  no  mention  be  made  of  interest. — 
Also: 

A  note  payable  on  demand,  will  draw  interest  after  a  demand  of 
payment  has  been  made,  for  it  is  then  due. 

9.  When  the  time  in  which  a  note  becomes  due  is  not  specified, 
it  becomes  due  as ;  oon  as  a  demand  of  payment  's  made. 

10.  A  1  notes  have  three  days  of  grace  after  they  are  nominally 
due,  before  they  are  legally  due. 

1 1 .  The  rate  of  interest  is  always  understood  to  be  the  le^al  rate, 
unless  otherwise  specified  in  the  note. 

12.  When  a  note  is  uiven  by  two  persons,  jointly  and  severally, 
^Note  3,)  it  may  be  collected  of  either  of  them,  but  not  of  both. 

13.  If  a  note  is  payable  by  installments,  the  amount  of  each  in- 
stallment, and  the  time  when  it  is  to  be  paid  should  be  specified  in 
the  note. 

14.  If  a  note  is  given  for  specific  articles,  as  wheat,  corn,  &c., 
payable  at  some  fixed  time,  and  paymnt  is  not  made  at  this  time, 
the  holder  can  claim  and  recover  the  value  of  the  note  in  money. 

Article  3.     Miscellaneous  Problems  and  Rules. 

Ex.  1.  A  vessel  6  feet  square  is  filled  with  water  to  the  depth  of 
10  feet,  what  pressure  does  the  bottom  sustain,  a  cubic  foot  of  wa- 
ter weighing  1000  ounces  Avoirdupois? 

^  6  X  6  =  36  sq.  ft.  area  of  the  bottom  of  the  vessel. 

Solution.    <  36  X  10  =  360  cubic  feet  of  pressure.  [Ans. 

(  360  X 1000  =  360000  oz.;  360000  -r-16=22500  lbs. 

Obs.  1.  The  sides  of  a  vessel  sustain  a  pressure  equal  to  ike 
area  of  the  sides  midtiplied  by  half  the  depth  of  the  water. 

2.  What  would  be  the  pressure  of  water  against  the  gates  of  a 
sluice  18  ft.  deep,  and  24  ft.  wide,  when  filled? 

-  .  Ans.  243000  lb«. 


Art.  3,  MISCELLANEOUS    PilOBLEMS    AND    RULES.  1  861 

3.  What  would  be  the  pressure  of  water  against  the  gates  of  a 
sluice  18  ft.  wide,  and  12  ft.  deep,  when  full?         Ans.  8 1000  lbs. 

To  find  the  height  of  an  object  by  knowing  its  distance: 

^   Obs.  2.     Tahe  the  distance  in  mUes :  tivo  thirds  of  the  sqiuire  of  this 
mil  he  the  height  of  the  object  y  in  feet. 

Note. — In  this  calculation,  of  course  there  must  be  no  obstruction  to  the 
range  of  vision  more  than  the  natural  convexity  of  the  earth. 

4.  Looking  across  a  plain,  I  saw  the  top  of  a  pole  exactly  10  miles 
distant.     Required  the  height  of  the  pole?  Ans.  G^  ft.  8  in. 

5,  I  saw  the  top  of  a  mast  which  I  knew  to  be  just  12  miles  dis- 
tant.    Required  the  height  of  the  mast?  Ans.  96  feet. 

To  find  the  distance  of  an  object  by  knowing  its  height : 

Obs.  3.  Take  the  height  in  feet;  increase  this  hy  one-half  of  itself 
and  extract  the  square  root  of  the  sum.  The  result  will  he  the  distance 
in  miles. 

6.  I  saw  the  top  of  a  mast,  which  I  knew  to  be  1 50  ft.  in  height. 
Required  its  distance?  Ans.   15  miles. 

7.  To  what  distance  on  a  level  plain  could  a  pole  be  seen  which 
measured  80  ft.  8  in,  in  height?  Ans.   1 1  miles. 

'To  find  the  time  a  body  has  been  falling: 

Obs,  4.     Divide  the  velocity  hy  32^-.     Or, 

Divide  the  space  reduced  to  feethy  IQj-^,  ond  extract  the  square  root 
of  the  quotient. 

8.  How  long  will  a  body  be  in  falling,  to  acquire  a  velocity  of  193 
feet  per  second?  Ans.  6  seconds. 

9.  I  dropped  a  stone  into  a  pit  400  ft.  deep.  How  long  was  it  in 
falling?  Ans,  4.98  + sec. 

10.  How  long  would  it  take  a  body  to  fall  2  miles? 

Ans.  25.62  + sec. 

To  find  the  velocity,  which  a  body  has  acquired  in  falling  : 

Obs.  6.     Multiply  the  space  reduced  to  feet  hy  16|-,  and  double  the 
square  root  of  the  product.     Or, 
Multiply  the  time  5y  32j.    • 

11.  A  stone  was  dropped  from  the  top  of  a  ledge  of  rocks  228 
ft.  in  height.  Required  its  velocity  per  second  when  it  had  reached 
the  bottom?  Ans.   121.11  ft. 

12.  What  velocity  per  second,  would  a  stone  acquire  in  falling  1 
mile?  Ans.  682.82  ft,  per  sec. 

17 


36^  COMMON    ARITHMETIC^  ScCt.    XVII 

13.  A  body  has  been  falling  12  seconds.  Required  the  velocity 
it  has  acquired?  Ans.  386  ft.  per  sec. 

To  find  the  space  a  body  has  fallen  throu^^h: 

Obs.  6.     Multiply  the  square  of  the  time  by  16y^.     Or, 
Divide  the  square  of  the  velocity  hy   64i-, 

14.  A  body  has  been  falling  9  seconds:  Required  the  space  it  has 
passed  through?         -  Ans.  1302 J  ft. 

15.  Wishing  to  ascertain  the  depth  of  a  chasm,  I  dropped  a  stone 
from  the  top  and  saw  it  strike  the  bottom  in  4  seconds.  Required 
the  depth  of  the  chasm? 

Ans.  257  ft.  4  in. 

16.  A  body  has  fallen  so  far  as  to  have  acquired  a  velocity  of 
579  ft.  per  second.     Through  what  space  has  it  passed? 

Ans.  5211  ft. 

Note. — In  the  above  rules,  no  allowance  is  made  for  the  resistance  of  the 
air. 

Given  the  base  (or  perpendicular)  of  a  right-angled  triangle,  to- 
gether with  the  sum  (or  difference)  of  the  other  two  sid  js,  to  find 
those  sides : 

Obs.  7.  To  the  square  of  the  given  side,  add  the  square  of  the  sum 
{or  difference)  of  the  other  two  sides,  and  divide  the  result  by  twice  this 
ium  {or  difference;)  the  quotient  will  he  the  length  of  the  longest  side. 

17.  The  base  of  a  right  angled  triangle  measures  16  ft.,  and  the 
sum  of  the  other  two  sides  is  32  ft.  Required  the  length  of  those 
sides?  Ans.  Hypoth.  20  ft.  Per.  12  ft. 

18.  The  foot  of  a  ladder  was  moved  out  12  ft.,  and  the  top  fell 
6  ft.  Required  the  length  of  the  ladder,  supposing  it  to  have  stood 
perpendicularly  against  a  wall  at  first?  Ans.   15  ft. 

19.  A  pole  90  ft.  high  was  broken  off  by  the  wind,  and  the  two 
parts  held  together.  The  top  fell  30  ft.  from  the  foot  of  the  pole. — 
Required  the  length  of  each  part?  Ans.  40  ft.  and  50  ft. 

To  measure  standing  timber: 

Obs.  8.  Place  two  stakes  so  that  they  will  range  with  the  top  of  the 
timber;  then  multiply  the  distance  from  the  short  stake  to  the  tree,  by  the 
difference  between  the  lenyth  of  the  stakes,  and  divide  the  product  hy 
the  distance  between  the  stakes.  The  residt  will  be  the  length  of  the  timber 
above  the  shorter  stake. 

20.  I  wish  to  find  a  stick  of  timber  36  ft.  long,  and  find  one  that 
has  a  knot  that  will  spoil  it  above  a  certain  htight.  I  therefore  place 
a  stake  10  ft.  long,  30  feet  from  the  tree,  and  6  ft.  from  this  I  place 


Art.  3.  MWCKLLANBOUS    PROBLEMS    AND    RULES.  363 

another  stake  4j  ft.  long,  so  that  the  top  of  the  two  stakes  wiil 
range  with  the  knot.  Will  the  stick  answer  my  purpose  by  allow- 
ing 2^  ft.  for  the  stump? 

Ans.  It  will  not  answer,  it  being  but  35  ft.  long. 

To  find  the  mean  temperature  of  any  day,  reckoning  from  sun- 
rise to  sunrise: 

'  Obs.  9.  Add  tog  ether  the  morning  observation,  twice  the  afternoon 
ohservation,  twice  the  evening  observation,  and  the  next  morning  obser- 
vation, and  divide  the  sum  by  6. 

Remark  1.  This  is  called  DeWitt's  Rule.  It  gives  the  correct  mean  tem- 
perature, on  the  supposition  that  the  thennometer  has  risen  and  fallen  regular- 
ly between  the  observations.  The  observations  are  taken  atsunrise,  at  2  P.  M., 
at  9  P.  M.,  and  at  sunrise  the  next  morning. 

2.  When  any  of  the  observations  are  below  zero,  that  quantity,  or  its  dou- 
ble, (as  the  case  may  me,)  must  be  subtracted,  instead  of  added. 

3.  The  mean  temperature  for  any  number  of  days  is  found  by  dividing  the 
sum  of  the  mean  temperature  of  all  the  days  by  their  number. 

21.  On  a  certain  day  the  thermometer  stood,  at  sunrise  38°  Fah- 
renheit, at  2  P.  M.  50°,  at  9  P,  M.  42°,  and  at  sunrise  the  next 
morning  30°.     Required  the  mean  temperature  of  that  day? 

Ans.  42°. 

22.  On  a  certaui  day  the  thermometer  stood  at  sunrise  — 6°,  at 
2  P.  M,  10°,  at  9  P.  M.,  4°,  and  at  sunrise  the  next  morning  at 
— 10°  ;  what  was  the  mean  temperature  of  that  day?    Ans.  2°. 

This  sign  ( — )  signifies  that  the  observation  was  below  zero. 

23.  A  square  field  is  to  be  enclosed  by  a  fence  8  rails  high,  and 
having  two  lengths  to  the  rod.  How  large  must  this  field  be 
that  it  may  contain  as  many  acres  as  there  are  rails  around  it,  or 
that  each  rail  may  fence  an  acre? 

Obs.  10.  Multiply  4  times  the  number  of  rails  on  one  rod  by  160  ; 
this  will  give  the  number  of  rods  on  one  side  of  the  field, 

Ans.  655360  acres  =  1024  sq.  miles  =  32  miles  sq. 

The  sum  and  product  of  two  numbers  being  given  to  find  the 
numbers: 

Obs.  1 1 .  From  the  square  of  half  their  sum  subtract  their  product, 
and  extract  the  square  root  of  the  remainder;  to  the  root  add  the  half 
sum,  and  the  result  will  be  the  greater  number. 

24.  The  sum  of  two  numbers  is  56,  and  their  product  768  ;  re- 
quired the  numbers?  Ans.  32  and  24. 

25.  A  man  bought  a  certain  number  of  acres  of  land  for  ^4375; 
if  the  number  of  dollars  he  paid  per  acre  were  added  to  the  num- 


S64  •GMVION    ARITHMETIC,  ScCt.  XVII 

ber  of  acres  bought,  the  sum  would  be  200.     How  many  acres  did 
he  buy,  and  what  did  he  pay  per  acre? 

Ans.  He  bought  175  acres,  at  ^25  per  acre. 

26.  Two  men,  A  and  B  bought  150  acres  of  land  for  $375,  of 
which  A  paid  $208,  and  B  $170;  on  account  of  a  difference  in  the 
quality  of  the  land, they  divided  it  so  that  A  paid  $1.20  .per  acre 
more  than  B.  How  many  acres  did  each  get,  and  what  did  he  pay 
per  acre? 

1        i  A  got  65  acres  at  $3.20  per  acre. 
(  B  got  85  acres  at  $2.00  per  acre. 

To  find  the  number  of  acres  obtained  by  him  who  paid  least^'per 
acre : 


Obs.  12.  Multiply  the  whole  number  of  acres  hy  the  difference  be- 
tween the  prices  per  acre,  subtract  the  product  from  the  price  paid  for  the 
whole  land,  and  divide  the  remainder  by  twice  the  difference  between  the 
prices  per  acre.  Square  the  quotient,  and  to  this  square  add  the  product 
of  the  whole  number  of  acres  multiplied  by  the  sum  paid  by  him  who 
paid  the  least  j^r  acre,  and  divided  by  the  difference  between  the  prices 
per  acre;  extract  the  square  root  of  the  sum,  and  from  this  root  subtract 
the  number  that  was  squared;  the  remainder  will  be  the  number  of  acres 
obtained  by  him  ivho  paid  the  least  per  acre. 

Solution  of  the  last  example.] 

[whole. 

150=  Ko,  acres.  $378    =    Amt.pd.  for  the 

$1.20=  Diff.  in  the  price  per  acre,       180    =    Prod,  to  be  sub- 

[tracted. 

180.00  Diff.   1.20  X  2  =  2.40)198.00(825  =  quotient  to 

1920  [be  squared. 

600 
480 

1 50  =  No.  acres.  1 200 

$170  =  Sum  paid  by  him  who  >  1200 

paid  least  per  acre.  ^  

105  0000 
15 


1.2iO}25500.0|0 


21250  =  Number  to  be  added    } 
to  the  square  of  82.5.  \ 
[Sclution  continued  on  next  pao^e.] 


Art.    3.  ^MISCELLANEOUS    PROBLEMS    AND    RULES.  363 


82.5 
r  82.5 

4125 
1650 
6600 

6806.25 
21250 

28056.25(167.5 
1 50-85  =65  =  Acres  obtained  by  A.     1  82.5=No.sqrd. 

$208-j-65=^3.20=price  per  acre )      

paid  by  A  ^  26)  1 80  85=  A'^res  ob- 

156  tained  by  B. 

327)2456 
$170-f-85  =  $2.00=  price  per  >  2289 


[acre  paid  by  B. 


3345)167  25 
167  25 


i^  000  00 

27.  A  and  B  bought  280  a-res  of  land  for  8738.  of  which  A 
paid  $326.25,  and  B  paid  $411.75.  The  land  was  divided  so  that 
B  paid  $0.80  per  acre  more  than  A*  How  much  land  did  each  get 
and  what  did  he  pay  per  acre? 

L        i  A  got  145  acres  at  $2.25  per  acre. 
(  B  got  135  acres  at  $3.05  per  acre. 
The  following  questions  are  solved  chiefly   by  analysis,    without 
reference  to  any  particular  rules*     They  are  easy  enough  to  be  un- 
derstood if  the  pupil  thoroughly  understands  the  principles  previous- 
ly explained. 

28.  A,  B,  and  C  agreed  to  divide  $50  between  them  giving  A  \, 

B  ~,  and  C  J.     What  was  the  share  of  each  in  this  proposition? 

SoltUion.—^  +  j  +  i-  =  Tl;  50-T-2  =25;  50-^3=  1 6|;  5 -h  4 
__  jgi 

Then  13:  12  :  $25  !  $231  A's  share  .13  ;  12  :  :  $16| :  $15y\ 
B's  share  .13:   12  :  I  $12^:  $11^3  C's  share. 

29.  $335  is  required  to  be  divided  between  A,  B  and  C,  in  such 
a  manner,  that  as  often  as  A  gets  $3  B  gets  $5,  and  C  gets  $7  as 
often  as  B  gets  ^4. 

Solution. — Since  C  gets  v7  as  often  as  B  gets  $4,  it  is  evident 
that  as  often  as  B  gets  ^5  C  gets  \  of  '^b  or  '4^\.  Hence,  their 
shares  are  in  the  proportion  of  3,  5  and  8 J;  or  12,  20,  and  35. 

-  Ans,  $60  A's  share.     $100  B's  share.     $475  C's  share. 

■  ■"  ♦ 


366  COMMON   ARITHMETIC.  Scct.    XVII 

30.  Suppose  A,  B,  and  C  start  at  the  same  point  and  travel  in  the 
same  direction  about  an  island  53  miles  in  compass,  A  traveling  at 
the  rate  of  3,  B  6,  and  C  7  miles  an  hour,  in  what  time  will  they 
next  come  together? 

Solution. — 5 — 3=  2  miles  B's  gain  per  hour  on  A. 
7—3=4     *'     C's         **         "     on  A. 
Dividing  the  distance  about  the  island  (53  miles)  by  the  greatest 
common  divisor  of  2  and  4,  the  gains  of  B  and  C  per  hour  on  A, 
we  find  that  they  will  all  meet  again  in  53  -f-  2  =26 ^  hours. 

Ans.  26^^  hours. 

31.  A  hare  is  60  leaps  before  a  greyhound,  and  takes  6  leaps 
while  the  greyhound  takes  5,  but  3  leaps  of  the  hound  are  equal  to 
4  of  the  hare.  How  many  leaps  must  the  greyhound  take  before 
he  catches  the  hare? 

SoliUion. — As  3  leaps  of  the  greyhound  are  equal  to  4  of  the 
hare,  5  leaps  of  the  greyhound  are  equal  to  |-  of  4  =  6|  leaps  of 
the  hare  ;  therefore  he  gains  |  of  a  leap  on  the  hare  for  every  5 
leaps  he  takes,  and  gains  1  leap  for  every  (5-r-|  =)  1\  leaps  he 
takes:  but  he  must  gain  60  leaps  to  overtake  the  hare,  therefore  he 
must  take  60  X  7^  =  450  leaps  to  overtake  her.  Ans.  450. 

32.  A  person  bought  a  certain  number  of  apples  at  the  rate  of 
two  for  a  cent,  and  afterwards  bought  as  many  more  at  the  rate  of 
3  for  a  cent:  he  sold  them  all  at  the  rate  of  5  for  2  cents,  and  lost 
2^  cents.     How  many  apples  did  he  buy  in  all? 

SoliUion. — 2  apples  for  1  cent  is  ^  a  cent  apiece ;  also: 

3  apples  for  1  cent  is  i-  of  a  cent  apiece.  Then  2  ap- 
ples (that  is  1  of  each  lot,)  cost  ^-|-i-=  J  of  a  cent,  and  1  apple 
cost  I -^2=  Y2  of  a  cent. 

Again,  as  he  sold  5  apples  for  2  cents,  he  sold  1  apple  for  2  -^-  5 
=1  of  a  cent.  Therefore  he  lost  y^ — |  =  ~  of  a  cent  on  each 
apple  he  sold;  and  1  cent  on  60  apples;  and  since  he  lost  2^  cents, 
he  must  have  sold  60X2}=  150  apples.  Ans.   150. 

33.  A,  B,  and  C  can  do  a  piece  of  work  in  5  days;  B,  C,  and  D 
in  6  days;  C,  D  and  A,  in  7  days;  and  D,  A  and  B,  in  8  days  ;  in 
what  time  would  they  all  do  it,  working  together,  and  in  what  time 
would  each  one  do  it  alone  ? 

Solution. — A,  B,  and  C,  can  do  ~  of  the  work  in  1  day. 

B,  C,  and  D  can  do  \      " 

C,  D,  and  A  can  do  \      *'         ^*         " 

D,  A,  and  B  can  do  ^      "         "         " 

Therefore  they  all  can  doj-|-i--f-y-j-J-  =  |-|J  of  the  work  in  3 
days,  or  II J  -^  3  =  ^y/o  of  it  in  1  day. 

Then  l-h  i'j%%  =  4||-|  days  in  which  they  all  would  it. 


Art,    3.  MISCELLANEOUS    PROBLEMS    AND    RULES.  867 

Also:  jjVo  —  i  =2lJo-  of  the  work  done  by  D  in  one  day,  and 
1  -r-  2  jJo=  ^^14  days,  in  which  D  would  do  it. 

In  the  same  manner  it  can  be  found  that  A  can  do  it  in  22y  j^ 
days;  B  can  do  it  in  Hy—  days;  and  C  can  do  it  in  llyVV  days. 

34.  A  certain  piece  of  work  is  to  be  performed,  requiring  a  fra- 
mer,  a  carpenter,  and  a  mason;  and  no  two  can  work  at  the  same 
time.  The  mason  receives  ^1.25  per  day;  the  carpenter  receives 
$2  per  day;  and  the  framer  receives  $2.50  per  day;  the  work  is  com- 
pleted in  100  days,  and  each  receives  the  same  amount  of  money. 
How  many  days  did  each  work,  and  how  much  did  each  receive? 

Solution, — As  the  framer  receives  ^2.50  per  day,  whilst  the  ma- 
son receives  but  ^1.25  per  day,  it  would  take  x'ii  =  ^  days  of  the 
mason  to  amount  to  1  of  the  framer's;  likewise,  it  would  take  |l^J 
=  1^-  days  of  the  carpenter  to  amount  to  1  of  the  framers.  There- 
fore, the  days  they  worked  are  to  each  other  as  1,  1  j,  and  2.  Then 
l  +  U+2=4l:;  and  4i:  I  1  :  :  100  ;  23fV  days  worked  by  the  fra- 
mer. 4^:  Ij  :  ;  100;  29yy  days  worked  by  the  carpenter. — 
And,  4f  :  2  :  I     100  I  47yY  days  worked  by  the  mason. 

And  23yV  X  2.50  =  .$58l-Y.) 

29^  X  2  =  $5S\^.        V  the  sum  received  by  each. 
47yV  X  1.25=^5814.    ) 


And  100  =  the  number  of  days  worked  by  all, 

35.  A  gentleman  died  leaving  a  son  and  daughter  in  foreign  coun- 
tries, and  directed  in  his  will,  that  if  the  son  only  returned,  he 
should  receive  |  of  the  estate  and  the  widow  the  balance;  but  if  the 
daughter  only  should  return,  the  widow  should  receive  f ,  and  the 
daughter  the  balance:  as  it  happened  both  returned,  and  thus  the 
widow  lost  in  equity  ^1600  dollars  more  than  she  would,  had  the 
daughter  only  returned.  How  much  was  the  share  of  each,  and 
how  much  would  the  widow  have  received  had  the  son  only  re- 
turned? 

Solution. — By  the  question,  the  son  receives  twice  as  much  as  the 
widow,  and  the  widow  twice  as  much  as  the  daughter;  therefore 
the  daughter  receives  1  part,  the  widow  2  parts,  and  the  son  4  parts, 
and  4  -j-  2  -j-l=:  7  parts  into  which  the  estate  is  to  be  divided. 

Had   the   daughter  only  returned,    the  widow  would  hare  re- 
ceived I  of  these  7  parts,  or  4|  parts  of  the  whole  estate;  but  both 
returning,  she  receives  4| — 2  =  2|  parts  less  than  she  would,  had 
the   daughter  only   returned.       But  these  2|  parts  =  $1600;    1 
1600 

part  therefore,  must  be =^600  ;  hence,  the  estate  is  $600X7 

21- 
=  $4200 ;  and  $4200  -h  3'==$1400,  the  widow's  share  had  the  son 


368  COMMON   ARITHMETIC.  Scct.    XVIII 

only  returned.     But  as  it  is  their  shares  are  to  each|other  as  the 
numbers  1,  2,  and  4;  then 
C7  :  1 


7:  2 
7:  4 


$4200  :  $600  the  daughter's  share. 
$4200:  $1200  the  widow's  share.     And 
$4200  :  $2400  the  son's  sh-are. 


SECTION  XVIII. 
MISCELLANEOUS  EXERCISES  FOR  THE  ;SLA.TE. 

1.  A  sum  of  money  is  to  be  divided  between  five  men;  A  re- 
ceives $130  ;  B  $170 ;  C  as  much  as  A  and  B  ;  D  as  much  as  B 
and  0 ;  and  E  as  much  as  all  the  rest.     Required  the  sum  divided? 

Ans.  $2080. 

2.  A  man  was  2 1  years  old  when  he  left  college  ;  he  studied  3 
years  for  a  profession;  then  traveled  15  years;  after  which  he  mar- 
ried and  lived  with  his  wife  30  years,  when  she  died,  leaving  him 
a  son  15  years  of  age  ;  the  son  was  26  years  old  when  the  father 
died.     How  old  was  the  father  at  his  death?         Ans.  80  years* 

3.  Add  together  4  quadrillions,  1 27  trillions,  9  billions,  1 8  millions, 
102  thousand  407;  39  trillions,  7  millions,  and  1;  18  trillions,  and 
17;  204  billions,  10  thousand;  7  millions,  4  thousands,  303 ;  and 
199.  Ans.  4184213032116927. 

4.  A  man's  property  was  worth  $75000:  he  lost  a  house  worth 
$3000  by  fire,  and  a  vessel  worth  $15000  in  a  storm.  Required 
the  value  of  the  remaining  property.  Ans.  $57000. 

5.  From  100  quadrillions,  100  billions,  and  1,  take  101  billions, 
101.  Ans.  99999998999999900. 

6.  A  farmer  sold  as  follows:  a  quantity  of  wheat  for  $125;  corn 
for  $200;  hay  for  $75;  oats  for  $50;  butter  and  cheese  $40;  and 
some  other  things  for  $25;  he  received  in  return  a  piece  of  "broad- 
cloth worth  $30;  $75  worth  of  satinet ;  $90  worth  of  groceries; 
$1.3  worth  of  muslin  ;  $40  worth  of  books,  and  the  balance  in 
money.     How  much  money  did  he  receive?  Ans.  $265. 

7.  Multiply  146  millions,  201  thousand.  111,  by  210  millions,  12 
thousand,  222.  Ans.  30704020179978642. 

8.  There  two  numbers;  the  lesser  number  is  1476  times  28921, 
and  their  differerence  is  1492  times  1728.     Required  their  product? 

Ans.  1932269397130512, 

9.  Divide  1  sextillion  by  333. 

Ans.  3OO3OO3OO3OO3OO3OO33I3. 

10.  Take  1  from  1  sextillion  and  divide  the  remainder  by  999. 

Ans.  1001001001001001001. 


MISCELLANEOUS    EXERriSES    FOR    THE    SLATE.  369 

11.  ITow  long  wo  il'l  i^  ^•"•Vo  foommt  a  trillion  by  counting  676(0 
per  day?  Ans.   1 7361 1 1 1  J-  days  =  47532  yrs.  48i  days. 

It.  ihe  bible  ct  ntiun.s  3i  1  73  viihes:  by  reading  74  verses  per 
day,  how  long  will  it  take  to  read  it  through? 

Ans*  421^1  days. 

13.  The  sum  of  two  numbers  is  18261,  and  one  of  the  numbers 
is  4927;  what  is  the  other  number!  Ans.   13334. 

14.  Two  men,  A  and  B  have  together  ^15000;  A  has  -^8976; 
bow  much  has  B?  Ans.  ^6024. 

15.  C  has  ^1000,  which  is  $496  dollars  more  than  D  has;  how 
much  has  D?  Ans.  .^504. 

16.  Two  men  together  live  150  cattle,  and  one  has  8  more  than 
the  other;  how  many  have  each?         Ans.  One  71;  the  other  79  v 

17.  What  number  multiplied  by  144  wih  produce  34272? 

Ans.  238. 

18.  What  number  divided  by  216  will  produce  383? 

Ans.    82728. 

19.  If  the  dividend  is  23188,  and  the  quotient  is  124,  what  is  the 
divisor?  Ans.   187 

20.  There  is  a  certain  number,  to  which  if  460  b^^  added,  and 
from  the  sum  3uu  be  subtracted,  and  the  remainder  divided  by^8, 
the  quotient  will  be  equal  to  9900  divided  by  132.  Wha  is  the 
number?  Ans.  440. 

21 .  A  lad  being  asked  Low  much  he  gave  for  a  book,  replied;  if 
you  11  vide  the  price  by  2,  to  the  quotient  add  1 1,  multiply  the  sum 
by  6  and  from  the  pro'luct  subtract  52,  the  remainder  \\ill  be  ^-^0 
Required  the  cost  of  the  book?  Ans.  $2. 

22.  A  peddler  made  three  trips;  the  second  trip  he  lost  ^25;  but 
the  third  trip  he  gained  $40;  in  the  three  trips  together  he  gained 
$25.     Did  he  gain  or  lose  the  first  tjip,  and  how  much? 

Ans.  He  gained  $10. 
1^23.  A  travels  30  miles  per  day,  and  B  40;  if  both  travel  the  same 
road,  and  A  has  5  days  the  start  of  B,  how  many  days  will  it  take 
B  to  overtake  A?  Ans  15. 

24.  Multiply  288  by  144,  216,  137.  2625,  1728  and  256,  and 
divide  the  product  by  576,  72,  274,  864,  1644,  135,  and  512. 

Ans.   IvVV. 

25.  A  owns  580  rods  of  land;  B  owns  796  rods;  C  owns  848 
and  D  ov^ns  1232  roJs  they  agree  to  divide  it  into  equal  lots  fixing 
on  the  greatest  number  of  rods  for  a  lot,  that  will  allow  each  owner 
to  lay  out  all  his  land,     How  many  rods  in  each  lot? 

y-:-^.  4. 

26.  A  iTT-entleman  hn>  ca-ks  lii/LHiu;  23,  36,  42,  arul  Go  oajjons 
respectively;  Avhat  is  the  smallest  number  of  gallons  that  will  jutt 
fill  same  number  of  casks  of  either  kind?  Ans.  252. 

17A 


370  COMMON     ARITHMETIC  Scct.    XVIII 


27.  What  fraction  is  that,  to  which  if  you  add  ~  the  sum  will  be  f 


Ans     ' 


1  5" 


28.  What  fraction  is  that,  from  which  if  you  subtract  y,  the  re- 
mainder will  be  4-  -A.ns.  —• 

29.  What  number  is  that  which  being  multiplied  by  12,  produces 

30.  What  number  is  that  which  being  multiplied  by  9,  produces 
*?  -  Ans.  Jj. 

31.  What  number  is  that  which  being  divided  by  6  produces  |? 

Ans,  4. 

32.  What  number  is  that  which  being  multiplied  by  |,  gives  J? 

Ans,   11, 

33.  What  number  is  that  which  being  divided  by  J  gives  \t 

Ans,  J. 

34.  What  number  is  that,  from  which,  if  you  take  j  of  itself,  the 
remainder  will  be  12?  Ans.  21. 

35.  What  number  is  that  to  which  if  you  add  |  of  f  of  itseif, 
the  sum  will  be  30?  Ans.  24, 

36.  What  number  is  that  to  which  if  you  add  |  and  J  of  itself 
the  sum  will  be  93?  Ans.  36, 

.37,  What  fraction,  is  that  to  which  if  you  add  |  and  j,  the  sum 
will  be  1?  Ans,  1. 

38.  ^  of  a  pole  is  in  the  air,  ^  of  it  in  the  water,  and  6  feet  is  in 
the  mud;  required  the  length  of  the  pole?  Ans.  28  feet. 

39.  J  of  a  certain  number  exceeds  ~  ot  it  by  4  ;  what  is  the 
number?  Ans,  48. 

40.  What  number  is  that  from  which  if  you  take^ §  of  J  of  itself 
and  to  the  remainder  and  |  of  |  of  itself,  the  sum  will  he  93|? 

Ans.   120, 

41.  What  number  is  that  to  which  if  12  be  added,  1  of  the  sum 
will  be  16?  Ans.  24, 

42.  A  young  man  spent  |  of  his  property  in  1 8  months,  and  |-  of 
the  remainder  in  12  months  after;  he  then]  had  $3000  left.  How 
much  was  his  fortune  at  first?  Ans,  $18000. 

43.  A  grocer  bought  a  number  of  boxes  of  tea,  each  containing 
56i-|  pounds,  paying  at  the  rate  of  $5  for  8  pounds,  and  sold  it  at 
the  rate  of  11  pounds  for  $8,  and  gained  $300,  How  many  boxes 
were  there?  Ans.  62, 

1  n  2^ 

44.  If  8  —  English  Guineas  are  worth  $44 and  16 —  Eng- 

n  16|  5 

lish  Sovereigns  are  worth  $QO~^j,  how  many  Sovereigns  are  worth 
1 

19 Guineas?  Ans.   20. 

101^ 

45.  If  A  does  as  much  work  in  \\  days,  as  B  dots  in  2|-    days, 


Miscellaneous  eXercisp:;s  for  tmr  slate. 


St7l 


ahd  C  does  asmiich  in  3|  days,  as  A  doe=!  in  2  J  days,  how   many 
days  work  of  B  are  equal  to  7^  days  work  of  C^  Ans.   11^, 

46.  A  younger  brother  received  ^1530,  which  was  just  ^^  of  his 
elder  brother's  fortune,  and  6|  times  the  elder  brother's  fortune  was 
|-  as  much  again  as  the  father  was  worth.  Required  the  value  of 
the  estate?  Ans.  ^19165f. 

47.  A  person  spent  all  his  income  and  ' 
but  after  that  he  saved  jV  of  his  income, 
his  deficiency  and  had  ^50  left. 

Ans.  ^1000. 

48.  What  number  multiplied  by  16  will  produce  .961 

Ans.  .06. 
produce  IS? 

Ans,  2,7. 

^5  equivalent?  Ans.  .0222-{-. 

What  is  the  value  of  the  decimal  .3625?  Ans.  |  J, 

In  return  he  received 
150  lbs.  sugar  at  $0.10  per  lb. 
201   lbs.  coffee   at   .08^     " 


49. 

60. 
61. 


What  number  divided  by 
To  what  decimal  is 


3  as  much  more  one  year, 
and  in  5  years  made  good 
How  much  was  his  income? 


.  5  wil 


ralent? 


62.  A  farmer  sold 
150  bu.  wheat  at  $1.12^perbu. 
300   bu.   corn    at    .37|^  per  bu. 
250   bu»   oats    at 
175  lbs.  butler  at 
412  lbs.  cheese  at 
1200  lbs  pork  at 


•31, 
.I2I 
.10 
,06^: 


per  bu. 
per  lb. 


30  galls  molasses  at  .44  per  gal. 

3  barrels  salt  at  1 .75  per  bl. 

2  quintals  fish  at  .06^  per  lb. 

30  yds.  cloth  at  5. 75  per  yd. 

24  yds.  satinet  at    1.50  per  yd. 

And  the  balance  in  money. 

How  much  money  did  he  receive?  Ans.  $224.75. 

53.  If  18  grains  of  silver  will  make  a  thimble,  and  12  pwts.  a 
teaspoon,  how  many  timbles  and  teaspoons,  of  each  an  equal  num- 
ber can  be  made  from  15  oz.  6  pwts.  of  silver?  Ans.  24. 

54.  Sound  moves  at  the  rate  of  1142  feet  in  a  second  of  time: 
now  I  saw  the  flash  of  a  cannon,  fired  from  an  eminence,  just  1  min- 
ute before  I  heard  the  report.     How  far  distant  was  the  cannon? 

Ans.   If  M.  7  fur.  32  rds.  4  yds. 

65.  I  saw  several  men  upon  an  eminence  shooting  at  the  rate  of 
20  shots  per  minute;  I  saw  the  flash  of  three  rifles  before  I  heard 
the  first  report.     How  far  distant  were  they? 

Ans.   1  M.  2  fur.  15  rds.  4  ft.  6  in. 

6Q.  The  atmosphere  presses  upon  all  surfaces  at  the  rate  of 
about  15  lbs,  to  the  square  inch:  how  much  pressure  then  is  sustain- 
ed by  a  surface  measuring  15  sq.  ft.?  Ans.  16  T.  4  cwt» 

57.  I  wish  to  cut  ofi^  just  1  acre  of  land  from  the  end  of  a  lot  22 
rods  wide.     How  long  a  piece  must  I  cut  off? 

Am.  7t\  rdd». 


"  ^ 


S72  COMMON    ARITHMETIC.  Sect.    XVIII 

58.  Borne  very  important  news  having  arrived  at  New  Yoik  71° 
1'  8"  W.  Ion,,  in  4  minutes  it  was  communicated  to  St.  Louis,  Mo., 
90°  15'  16"  by  magnetic  telegraph.  At  what  time  was  it  known  in 
St.  Louis,  it  being  known  in  N.  Y.  at  half  past  2  o'clock,  P.  M.? 

Ans.  29  min.  4  sec.  past  1  o'clock,  P.  M. 

59.  A^traveler  who  had  set  a  perfectly  accurate  watch  by  the 
sun  in  Augusta,  Me.,  69°  50'  W.  Ion.,  being  in  Cincinnati,  84°  27', 
a  few  days  after,  was  surprised  to  find  it  wrong  when  compared  with 
the  sun.      Was  it  too  fast  or  too  slow?     How  much,  and  why? 

Ans.  It  was  58  min.  28  sec.  too  fast,  because  the  time  is  earlier 
at  Augusta  than  at  Cincinnati. 

60.  If  a  man  travel  376  miles  in  1 8  days,  how  far  can  he  travel 
in  27  days?  Ans.  564  miles. 

Note.     This,  and  the  47  following  examples  are  to  be  solved  by  Analysis. 

61.  If  838.25  pay  for  17  barrels  of  salt,  how  many  barrels  may 
be  bought  for  824.75?  Ans.   11. 

62.  it  a  barrel  of  flour  lasts  17  persons  5~  weeks,  how  long  will 
i  ^last  23  persons?  A.ns.  4~  weeks. 

63.  If  16  J  acres  of  land  cosrs  $325|,  what  will  26^  acres  cost? 

Ans,  .'^523,806. 

64.  Alias  120  acres  of  land;  |  of  A's  is  equal  to  yj-  of  B's. 
How  many  acres  has  B?  Ans.   165. 

65.  A  m.an  owning  ^  of  a  store,  sold  \j  of  his  share  for  '1i>2928; 
required  the  worth  of  the  store?  Ans.  ^5807.20. 

66.  Two  expresses  start  :.t  the  same  time  from  two  places  250 
miles  apart,  ;ind  ride  towards  each  other;  A  rides  Q\  miles  per  hour, 
and  B  rides  1 1|  miles  ;  er  hour.  How  far  does  each  ride  before 
they  meet? 

Ans.  A  rides  105y%\  miles:  B  rides  144i-|f  miles. 

67.  A  man-of-war  in  pursuit  of  a  smuggler  sails  8  miles  whilst 
the  smuggler  sails  5;  the  ship  passes  a  certain  island  when  the  smug- 
gler is  24  miles  beyond?  How  far  must  the  man-of-war  sail  to 
overtake  the  other?  Ans.  64  miles. 

68.  '*As  I  was  hunting  on  the  forest  grounds, 
Up  starts  a  hare  before  my  two  greyhounds ; 
The  dogs  being  light  of  foot  did  fairly  run. 
Unto  her  fifteen  rod?,  just  twenty-one  ; 

The  distance  that  she  started  up  before. 

Was  four  score  sixteen  rods,  just,  and  no  more. 

'Now  this  I'd  have  yon  unto  me  declare. 

How  far  they  ran  before  they  caught  the  hare." 

Ans.  336  rds. 

69.  A  set  out  from  Boston  for  Hartford  precisely  at  the  time  when 
B  at  Hartford   set   out  for  Boston,  distant  100  miles;  after  seven 


MISCELLANEOUS    EXERCISES    FOR  THE    SLATE.  373 

hours  they  met  when  it  appeared  that  A  had  traveled  1^  miles  per 
hour  more  than  B.     At  what  rate  per  hour  did  each  travel? 

Ans.  A,  7||  miles,  and  B  6  ^^  miles  per  hour. 

70.  A  cistern  has  two  pipes,  one  of  which  will  fill  it  in  24  hours, 
agdthe  other  in  30  hours;  it  has  also  a  dischaiging  pipe  which  will 
empty  in  18  hours.  If  all  are  left  open,  how  long  will  it  take  the 
cistern  to  fill?  Ans.  5\'~  hours. 

71.  xA  lion  of  bronze,  placed  upon  the  basin  of  a  fountain,  can 
spout  water  into  the  basin  through  his  throat,  his  eyes,  and  his  right 
f(  ot.  If  he  spouts  through  his  throat  only,  he  will  fill  the  basin  in 
6  hours;  if  through  his  right  eye  only,  he  will  fill  it  in  2  days  ;  if 
through  his  left  eye  only,  in  3  days;  and  if  through  his  foot  only, 
he  will  fill  it  in  4  hours.  In  what  time  will  the  basin  be  filled  if  the 
water  flow  through  all  the  apertures  at  once? 

Ans.  2^  J  hours. 

72.  A  can  mow  10  acres  in  6  days  ;  B  can  mow  13  acres  in  9 
days;  and  C  can  mow  17  acres  in  12  days:  in  what  time  can  they 
all  working  together  mow  108|  acres?  Ans,   24  days. 

73.  A  man  left  Si 3000  to  be  divided  between  his  wife,  son,  and 
daughter;  he  left  -Si 500  more  to  the  son  than  to  the  daughter,  a 
$2500  more  to  his  wife  than  to  his  son.  Required  the  share  o" 
each? 

Ans.  Wife's  share  ^$6500;  Son's  share  84000;  Daughter's  share 
$2500. 

74.  Thomas  sold  150  pine  apples  at  $0.33^-  apiece,  and  received 
no  more  money  than  Harry  did  for  a  certain  Jnuraber  of  watermel- 
lons  at  $0.25  apiece.  How  much  money  did  each  receive,  and  how 
many  melons  had  Harry? 

Ans.  Each  receives  $50,  and  Harry  had  200  melons. 

75.  A  farmer  sold  50  bushels  of  wheat,  at  $1.06^-  per  bushel,  and 
took  his  pay  in  cloth  at  $2.37^  per  yard.  How  many  yards  did  he 
receive?  Ans.  22--^. 

76.  A  goldsmith  mixed  3  oz.  of-  silver  14  carats  fine,  with  4  o^- 
17  carats  fine,  5  oz.  20  carats  fine,  and  6  oz.  23  carats  fine.  Re" 
quired  the  fineness  of  the  mixture?  Ans.   19-  carats. 

77.  A's  age  is  double  B's,  and  C's  is  3  times  A's;  the  s  mi  of  all 
their  ages  is  135  years.     Required  the  age  of  each? 

Ans.  A's  age  30  yrs.:  B's  15  yrs.;  C's  90  yrs. 

78.  A  man  bought  a  chaise,  horse  and  harness  for  $300;  the 
horse  cost  twice  as  much  as  the  harness,  and  the  chaise  cost  3  times 
as  the  horse  and  harness  togrether.     Required  the  cost  of  each? 

Ans.  Harness  $25;  horse  $50;  chaise  $225. 

79.  A  man  has  $24  to  pay  20  laborers;  he  pays  each  boy  3  cents, 
as  often  as  he  pays  each  woman  6  cents,  and  each  man  7  cents;  for 


St4  COMMON    ARITflMEtlO.  Sect.    XVlII 


every  boy  there  were  3  women,  and  for  every  woman  2  men.  How 
many  were  there  of  each,  ani  how  much  did  each  receive? 

C  There  were  2  boys,  6  women,  and  12  men. 
Ans.  <  The  boys  received  $0.60  each-  the  women  $1  each;    and 

^      the  men  ^1.40  each. 

80.  If  18  workmen  do  a  piece  of  work  in  20  days,  working  10 
hours  per  day,  how  many  workmen  will  it  take  to  do  tiie  same  work 
in  15  days  working  12  hours  per  day?  Ans*  20. 

81.  In  an  orchard,  ~  of  the  trees  bear  apples,  \  bear  peaches,  \ 
bear  plums,  and  30  bear  cherries.     How  many  trees  in  the  orchard? 

Ans.   120, 

82.  A  daughter's  portion  is  J  of  a  son's  portion;  both  their  por- 
tions amount  to  ^7200;  what  is  the  portion  of  each? 

Ans.  Son's  portion  $4000;  Daughter's  portion  $3200. 

83.  What  sum  of  money  is  that  whose  ~,  ~  and  j  parts  together 
amount  to  $141?  ^  Ans.  $180. 

84.  Wh  .t  number  is  that  to  which  if  25  be  added,  -q-  of  the  sum 
will  be  21?  Ans.  46. 

85.  What  number  is  that  which  increased  by  ~  of  itself,  is  equal 
to  8  of  itself  increased  by  9?  Ans.  24. 

86.  Out  of  a  cask  of  wine,  from  which  |-  part  had  leaked  21  gal- 
lons were  drawn,  when  the  cask  was  found  to  be  half  full.  How 
much  did  the  cask  hold?  Ans.   126  gallons. 

87*  A  person  having  spent  I  of  his  incolne,  found  he  wanted 
$25  of  having  |  of  it  remaining.     Required  his  income? 

Ans.  $600. 

88.  A  man  bought  a  quantity  of  beef  for  $7;  he  used  25  lbs.,  and 
then  sold  ~  of  the  remainder  for  $1.20  which  was  just  what  it  cost. 
How  many  pounds  did  he  buy  at  first?  Ans.   175. 

89.  A  father  divided  his  estate  between  his  children,  giving  the 
first  i  wanting  $300,  the  second  ~  wanting  $150.  and  to  the  third 
the  balance,  which  was  \  wantmg  $50.     How  much  was  the  estate? 

Ans.  $6000. 

90.  A  general  after  a  battle  found  that  he  had  left  500  men  more 
than  I  of  the  whole  army  fit  for  action;  250  more  than^f  the  whole 
being  Wounded;  125  more  than  -  of  the  whole  being  killed;  and  the 
remainder,  which  was  -j  of  the  Whole,  missing.  Of  how  many 
men  did  the  army  consist  at  first?  Ans.   15000. 

91.  A  hare  starts  12  rods  before  a  greyhound,  but  is  not  per- 
ceived by  him  until  she  has  been  up  l\  minutes;  she  scuds  away  at 
the  rate  of  36  rods  a  minute,  and  the  dog  in  view  makes  after  at 
the  rate  of  40  rods  a  minute.  How  long  will  the  course  last,  and 
how  far  will  the  dog  run? 

Ans.  The  course  lasts  14^  minutes,  and  the  dog  runs  570  rods. 

92.  A  and  B  have  the  same  income;  A  contr.icts  an  annual  debt 
equal   jj  of  hisi,  but  B  lives   on  |  of  his,  aTid   in  8  years    lends   A 


MISCELLANEOUS    EXERCISES    FOR    THE    SLATK,  ^Ifb 

enough  to  pay  his  debts,  and  has  S200to  spare.     How  much  is  the 
income  of  each?  Ans.  ^300. 

93.  A  man  being  asked  the  time  of  day,  answered  that  the  time 
past  noon  was  equal  to  5  of  the  time  till  midnight;  what  time  was  it? 

Ans.  30  minutes  past  4  o'clock.. 

94.  A  man  performing  a  journey  found  that  the  distance  he  had 
traveled  was  equal  to  y  of  the  distance  he  yet  had  to  go;  how  far 
had  he  traveled,  and  how  long  was  his  journey  in  all? 

Ans.  His  journey  was  165  miles,  and  he  had  traveled  60  miles. 

95.  Three  men  have  a  certain  sum  of  money  ;  A  has  '^200;  B 
has  as  much  as  A,  and  ^  as  much  as  C;  and  C  has  as  much  as  A  and 
B  both.     How  much  have  they  all  together?  Ans.  ^1400. 

96.  A  lady  has  two  silver  cups  and  only  one  cover,  which  weighs 
6  ounces:  if  the  cover  be  put  on  the  first  cup  its  weight  will  be  dou- 
ble that  of  the  second  cup;  but  if  the  cover  be  put  on  the  second 
cup;  its  weight  will  be  j  of  the  first  cup.  Required — the  weight  of 
each  cup? 

Ans.  Weight  of  the  first  cup  25  oz.;  of  the  second  cup  15  oz. 

97.  At  what  time  between  5  and  6  are  the  hands  of  a  clock  to- 
gether. Ans.  27  mm.  'iQj\  sec.  past  5. 

9^.  A  harmless  dove  was  soaring  high, 

To  stretch  her  wings  in  space  ^ 
At  length  a  hawk  did  her  espy, 

And  gave  the  dove  a  chase. 
Just  forty  rods  was  there  between 

These  birds,  as  we  could  view — ■- 
And  whilst  the  dove  flew  seventeen. 

The  hawk  flew  twenty-two. 
The  hawk  pursued  with  all  his  strength, 

As  those  who  saw  did  say  , 
Then  tell  the  rods  he  flew  in  length, 

Before  he  caught  his  prey. 

Ans.   176  rds. 

99.  At  what  time  between  8  and  9  are  the  hands  of  a  clock  ( x^ 
actly  opposite  each  other?  Ans.   10  min.  54yy  sec.  past  8. 

100.  A  man  bought  a  lot  of  wheat,  and  gave  \  of  it  to  some  poor 
families,  and  then  sold  the  remainder  for  $1 .  33^^  per  bushel,  and  re- 
ceived what  the  whole  cost  him.     How  much  did  it  cost  per  bushel? 

Ans.  Si. 00. 

101.  What  number  is  that  from  which  if  you  take  J  of  J  of  it- 
self, and  to  the  remainder  add  J  of  ^%-  of  the  number,  the  sum  will 
be  50?  "  Ans.    120, 

102.  A  man  when  he  married  was  3  times  as  old  as  his  wife  ;  15 
years  after  he  was  but  twice  as  old  as  his  wife.  Required — t lie  age 
of  each  when  married? 

Ans,  Man's  age,  45  yr^: ;  Wife's  age,  15  yrs. 


376  COMMON    ARITHMETIC,  ScCt.    XV II I 

103.  Divide  #450  between  A,  B,  and  C,  giving  A  ^50  more, 
and  B  $50  less  than  C. 

Ans.    A  8200;  B   flOO;    and   C  $150. 

104.  There  are  two  numiiers  whose  difference  is  8;  ~  of  the  less 
is  equal  to  3  of  the  greater.     Required — the  numbers? 

Ans.  40  and  48. 

105.  If  one  ship  containing  150  hogsheads'  of  wine,  pays  for  toll, 
at  the  Sound,  the  value  of  two  hogsheads,  wanting  $6,  and  another 
ship  containing  240  hogsheads  pays  at  the  same  rate  the  value 
of  two  hogsheads,  and  $18  besides,  what  is  the  value  of  the  wine 
per  hogshead?  Ans.  i23. 

106.  The  smaller  of  two  numbers  is  15,  and  if  we  add  to  this  \ 
of  both  numbers,  the  sum  will  be  the  greater  number.  Required — 
the  greater  n  imb(!r?  Ans.  25. 

107.  A  man  having  $50,  spent  a  part  of  it;  he  afterwards  re- 
ceived 6  times  as  much  as  he  had  spent,  and  then  had  twice  as  much 
as  he  had  at  first.     How  much  did  he  spend?  Ans.  $10. 

108.  If  2  men  can  do  a  piece  of  work  in  25  days,  and  it  takes  3 
women  the  same  time  to  do  it,  in  what  time  will  one  man  and  one 
woman  together  perform  it?  Ans.  30  days. 

109.  A  man  left  his  daughter  jj  of  his  property,  to  his  wife  jj  of 
the  remainder,  and  to  his  son  what  was  left.  Required — the  share 
of  each,  the  shares  of  the  sou  and  daughter  together  amounting  to 
$4500. 

.         i    Daughter's  share  $2000,  Wife's    share  $3000;  Son's 
^^^-  I  share  $2500. 

110.  If  cloth  worth  $0.25  cash,  is  worth  $0.31i  in  trade,  what 
is  wheat  worth  in  trade  that  is  worth  $1 .06^  cash? 

Ans.  $1.32l|-. 

111.  At  the  rate  of  $72  for  64  days'  work,  how  many  days  must 
a  man  work  to  earn  $99?  Ans.  88. 

112.  If  the  hind  wheel  of  a  wagon  is  14  ff.  8  in  circmuference,  and 
the  forward  wheel  11  ft.  9  in,  in  circumference,  how  many  times 
will  the  forward  wheel  turn  over  more  than  the  hind  wheel,  in  going 
470  miles?  Ans.  42000  times. 

113.  If  I  of  a  farm  is  worth  $2100,  what  is  -J^  of  it  worth? 

Ans.   1400. 

114.  If  5ij  acres  of  land  are  worth  nB34^  what  are  9~  acres 
worth?  Ans.    $60.20. 

115.  If  \5  men  mow  250  acres  of  grass  in  10  days,  working  12 
hours  per  day,  how  many  men  wiil  it  take  to  mow  300  acres  in  6 
days,  v/orkin  ;  10  hours  per  day?  Ans.  36. 

116.  If  33  men,  worki  g  10  hours  per  day,  dig  a  ditch  450  yards 
long,  5  feet  wndo,  and  4  feet  deep,  in  15  dn.ys  how  many  days  will 
it  take  45  men  to  dig  a  ditch  540  yards  long,  4  feet  wide,  and  3  feet 
deep,  working  1 1  hours  per  day,  if  the  strength  of  the  former  party 


11  MISCELLANEOUS    EXAMPLES     FOR   THE    SLATE.  377 

is  only  J  of  that  of  the  latter,  and  the  hardness  of  the  ground  in  the 
latter  case  is  y  times  that  in  the  former?  Ans.  6  days. 

Note. — Let  this  question  also  be  solved  by  Analysis. 

117.  A,  B,  C,  and  D  trade  together :  A  puts  in  i-,  B  J^.  C  i,  and 
D  the  rest;  they  gain  ^5400.     What  is  each  one's  share?  v   J 

Ans.  A's^i^lSOO;  B's  ^1350;  C's  ii^90U;  and  D's  81350. 

118.  If  the  third  of  6  were  3, 

What  would  the  fourth  of  20  be?        Ans*  7^. 

119.  If  6  and  4  just  9  had  been,] 

;Pray  tell  how  much  were  7  and   10? 

Ans*    15yu-. 

120.  A  bankrupt  owesas  follows  :  to  A  $1000;  to  B  $1200;  to 
C.  $800;  and  to  D  $1600;  his  property  is  worth  $4000.  How 
much  mus   each  creditor  receive? 

.        i    A,  $888.8881;  B,  $1066. 666|;  C,  $lll.lllj-; 
^^^'  I  D,  $1333. 333i. 

121.  In  a  storm  at  sea,  a  vessel  loaded  with  flour  was  obliged  to 
throw  overboard  150  barrels;  of  the  cargo,  A  owned  1000  barrels; 
B  owned  1400;  C  owned  1600;  and  D  owned  2000.  What  part  of 
the  loss  must  each  sustain? 

Ans,  A,  25  bis.;  B,  35  bis.;  C,  40  bis.;  and  D,  50  bis. 

122.  A  prize  of  $400  is  to  be  divided  between  a  Captain ,  two  lieu- 
tenants, and  6  soldiers  :  the  Captain  receives  a  share  and  a  half,  the 
lieutenants  a  share  each,  and  the  soldiers  a  half  share  each. — 
How  much  does  each  receive? 

Ans.   Captain,  $92  j\;  Lieut.  $61y\  each;  Soldiers,  $30 V|. 

123.  A,  B,  C,  and  D  agree  to  trade  together  with  $1200  capital, 
of  which  A  was  to  furnish  ^,  B  J,  C  j,  and  D  |;  D  withdraws,  and 
C  dies.  What  part  of  the  stock  must  A  and  B  furnish,  to  have  the 
same  proportion  as  before?  Ans.  A,  $686y;  B,  $5147. 

124.  A  commenced  trading  with  $1000  capital;  at  the  end  of  3 
months  betook  in  B,  with  $1400  ;  two  months  following,  they  took 
in  C,  with  $1200;  and  4  months  after,  they  took  in  D,  with  $1500: 
at  the  end  of  the  year  they  find  they  have  gained  $800.  How 
much  is  each  one's  share? 

Ans.  A's,  $256;  B's,  $268.80;  C's,  $179.20;  and  D's,  $96. 

125.  A  hired  two  horses  and  a  carriage  to  go  30  miles  and  back, 
for  $20:  after  proceeding  12  miles,  he  took  in  B  to  ride  through  and 
back  again  ;  when  within  5  miles  of  the  end  of  their  journey,  they 
took  in  C,  who  also  rode  back  with  them;  and  when  half  way  back, 
they  took  in  D,  who  rode  the  balance  of  the  way.  How  much 
ought  each  to  pay? 

Ans.  A,  $7.50;  B.  $6.25,   C,  $4.37^;  andD,  $1.87^. 

126.  Sold,  $1500  worth  of  goods  for  a  friend.  How  much  did 
my  commission  amount  to,  at  I2j  per  cent?  Ans.  $187.60 


378  COMMON    ARITHMETIC.  ScCt.    X  VIII 

127.  "What  will  be  the  cost  of  insuring  $2500  worth  of  property, 
at  3|- per  cent?  Ans.  $90. 62^. 

128.  A  merchant  bought  a  lot  of  goods  for  $21500,  and  paid 
^150  for  transportation.  For  how  much  must  he  sell  them  to  gain 
18  per  cent?  Ans.   $25547. 

129.  Bought  a  quantity  of  goods  on  credit  for  $75,  but  for 
cash  12  per  cent  was  deducted.     How  much  did  they  cost? 

Ans.  $66, 

130.  Bought  goods  for  $1200,  and  sold  them  for  $1400.  What 
per  cent  was  gained?  Ans.   16|. 

131.  Sold  tea  for  $1.12^  per  pound,  by  which  28y  per  cent  was 
gained.     Required — its  cost?  Ans,  $0.87|-. 

132.  Bought  goods  for  $500,  and  sold  them  for  $450.  What  per 
cent  was  lost?  Ans.   10. 

133.  What  is  the  interest  and  amount  of  $180  for  2  yrs.  4  mo. 
23  da.  at  6  per  cent?  Ans.  Int.  $25.89;  Amt.  $205.89. 

134.  What  is  the  interest  and  araoun'  of  $360  for  3  yrs.  5  mo. 
29  da.  at  7  per  cent?  Ans.  Int.  $88. 13;  Amt.  $448. 13. 

135.  A  man  gave  $300  for  the  use  of  $2500,  2  years.  What 
was  the  rate  per  cent  allowed?  x\ns.  6. 

136.  A  gentleman  gave  $4500  for  $3800  which  he  had  used  3 
years  6  mo.  15  da.  What  rate  per  cent  did  he  allow?    Ans.  5-\-. 

137.  In  what  time  would  $300  amount  to  $408,  at  6  per  cent? 

Ans.  6  years. 

138.  What  principal  would  amount  to  $221.40  in  1  yr.  9  mo.  12 
da.,  at  6  per  cent?  Ans.  $200. 

139.  What  sum  w .uld  gain  $55.30  in  2  yrs,  7  mo.  18  da.,  at  6 
per  cent?  Ans.  $350. 

140.  1  have  a  note  of  $540.50,  due  me  in  2  yrs.  6  mo.  without 
interest :  my  creditor  wishes  to  pay  me  now.  What  discount  ought 
I  to  make?  Ans.  $70.50. 

141.  A  certain  person  wished  to  sell  his  property,  and  a  gentle- 
man offered  him  $2100  for  it  at  the  present  time,  or  $2350  payable 
in  2  yrs.  11  mo;  he  chose  the  latter.  Did  he  gain  or  lose,  and  how 
much,  the  money  being  worth  6  per  cent?        Ans.  He  lost  $100. 

142.  What  is  the  difference  between  the  interest  of  $1000  for  3 
yrs.  7  mo.  6  da.,  at  7  per  cent,  and  the  discount  of  the  same  sum, 
for  the  same  time,  and  at  the  same  rate?  Ans.  50.722. 

143.  What  is  the  present  worth  of  6  per  cent  of  $2400.  payable, 
1^  in  6  mo.,  |-  in  9  mo.,  and  the  rest  in  1  year.     The  discount? 

Ans.  Present  worth,  $2296.965;  Discount",  $103,035. 

144.  A  broker  subscribed  for  40  shares  in  a  bank,  each  share  be- 
ing $100:  he  paid  in  40  per  cent  of  his  stock  when  he  subscribed, 
50  per  cent  of  the  remainde;  4  months  after,  and  the  balance  3 
months  following.  In  12  months  after  he  subscribed  there  was  a 
dividend  of  4j  per  cent  of  the  st-ock  among  the  stockholders,  and 


II  MISCELLANEOUS    EXERCISES    FOR   THB    BLATS.  879 

a  dividend  of  3  per  cent  accrued  each  6  months  afterwards.  At  the 
end  of  4  yrs.  6  mo.  the  broker  sold  his  stock  at  8i-  per  cent  advance. 
Now  supposing  he  hired  his  money,  and  paid  up  his  notes  when  he 
sold  his  stock,  did  he  gain  or  lose  by  the  speculation,  and  how  rauchl 

Ans.  He  gained  ^401.60. 

145.  A  certain  room  measures  20  ft  10  in.  in  length,  16  ft.  in 
width,  and  7  ft.  10  in.  in  height:  now  deducting  two  doors,  each 
measuring  6  ft.  7  in,  by  3  ft.  6  in.,  and  three  windows,  each  mea- 
suring 6  ft.  4  in.  by  3  ft.  4  in.,  what  will  it  cost  to  plaster  this 
room  at  $0.20  per  square  yard?  Ans.  '$11. 82 J. 

146.  A  certain  brick  building  measures  64  ft.  9  in.  in  length,  40 
ft.  8  in.  ii)%idth,  and  30  ft.  6  in.  in  heighth,  and  the  walls  are  1  ft, 
thick.  It  has  two  partition  walls  through  the  length  of  the  building, 
and  6  additional  walls,  each  15  ft.  2  in.  in  length;  fr^illf  the  side  of 
the  building  to  these  partitions,  all  of  the  same  heighth  and  thick- 
ness as  the  outside  walls.  Now  deducting  two  outside  doors, 
each  measuring  7  ft.  8  in.  by  3  ft.  6  in.  and  24  inside  doors,  each 
measuring  6  ft.  9  in.  by  2  ft.  8  in.,  and  36  windows,  each  measuring 
6  ft,  8  in.  by  3  ft.  9  in.,  how  many  bricks  did  it  require  for  this 
building,  each  brick  being  8  in.  in  length,  4  in.  in  width,  and  2  in, 
in  thickness,  and  how  much  did  they  cost  at  $3.50  per  M.? 

Ans.  It  required  31 1.202  bricks,  and  they  cost  $1089.207. 

147.  What  number  added  to  /j  of  123,  is  equal  to  ~j  of  the 
square  of  64?  Ans.   1120. 

148.  What  number  is  that,  which  substracted  from  6  times  the 
square  root  of  1849,  leaves  >-  of  552?  Ans.   120. 

149.  A  company  of  men  spent  $51.84,  each  man  spending  4 
times  as  many  cents  as  there  w.^re  men  in  the  company.  How  many 
men  were  there,  and  how  mu'ih  did  each  spend? 

Ans.  36  men;  each  spent  $1.44. 

150.  A  General,  forming  his  army  into  a  square,  found  he  had  76 
men  over  and  above  a  square;  but  by  increasing  each  side  by  1  man, 
he  wanted  161  men  to  complete  the  square.  How  many  men  had 
he?  Ans.   14000. 

151.  If  A  travels  due  north  24  miles,  and  B  due  east  32  miles, 
how  far  apart  are  they?  Ans,  40  miles. 

152.  A  certain  triangle  measures  16  feet  on  each  side.  Required 
— the  length  of  a  perpendicular  from  any  angle  to  to  the  opposite 
side?  Ans.   13.85-j- feet. 

153.  If  a  circle  measures  9  inches  in  diameter,  what  is  the  diam- 
eter of  one  4  times  as  large?  Ans.   18  inches, 

154.  If  the  circumference  of  a  circle  is  36  feet,  what  is  the  cir- 
cumference of  one  that  contains  but  -J  as  much?  Ans.   12  ft. 

155.  The  distance  between  two  places  is  such,  that  if  you  increase 
its  cube  by  16,  from  the  sum  subtract  30,  and  multiply  the  remain- 
der by  2,  the  product  will  be  100.     Required — the  distance? 

'  'i  Ans.   4  miles. 


380 


COMMON    ARITHMETC. 


Sect.  XVIII 


Ans,  ^ 


i  156.  In'measuring  a  certain  figure  like 
the  one  in  the  margin,  it  was  found  that 
the  length  of  the  lines  were  such,  that  as 
often  as  8  inches  were  measured  from  A 
to  B,  15  inches  were  measured  from  A  to 
E;  and  as  often  as  3  inches  were  measured 
from  A  to  E,  2  inches  were  measured  from 
A  to  C;  and  as  often  as  6  inches  were 
measured  from  A  to  D,  5  inches  were 
measured  from  A  to  C;  the  length  of  the 
four  lines  is  675  inches.  Required — the 
length  of  each  line,  and  their  distance 
apart  at  their  extremities?  B 

'  Distance  from  A  to  B  120  inches, 
I*         *'     A  to  C  150  inches. 
''     A  to  D  180  inches. 
*'     A  to  E  225  inches. 
"     B  to  C    90  inches. 
•*         **     B  to  D  216.333+ inches. 
*'     B  to  E  312, 129-f  inches. 
"     C  to  D  150  inches. 
"     C  to  E  270.416+  inches. 
"         ''     D  to  E  135  inches. 

157.  There  are  two  "columns  in  the  ruins  of  Persepolis  left  stand- 
ing  upright;  one  is  64  feet  high,  and  the  other  50  feet.  In  a  direct 
line  between  these,  stands  an  ancient  statue,  the  head  of  which  is 
97  feet  from  the  top  of  the  higher  column,  and  86  feet  from  the  top 
of  the  lower  column.  The  distance  from  the  foot  of  the  lower  col- 
umn to  the  foot  of  the  statue  is  76  feet.  Required — the  distince 
between  the  tops^of  the  columns? 

.        i  169.96  ft.  nearly.     Or, 
^'  I  157,04  ft.  nearly. 

Note. — The  learner  will  perceive  that  this  question  admits  of  two  difFetent 
answers,  according  ^as  we  suppose  the  statue  to  be  higher  or  lower  than  the 
columns. 

158.  How  many  cubical  blocks,  each  measuring^  j  of  an  inch  on 
a  side,  will  it  take  to  till  a  cubical  box,  each  side  of  which  is  2  feet? 

Ans,   4096. 

159.  In  a  cubical  foot,  how  many  cubes  measuring  2  inches  on 
each  side?  Ans.  216. 

160.  What  is  the  difference  between  a  solid  half  foot,  and  half  a 
solid  foot?  Ans.  648  s.  in. 

161.  Required — the  length  of  one  side  of  a  solid  body  contaii.ing 
\  as  much  as  another  solia  body  measnring  6  inches  on  each  side? 

Ans,  3  inches. 


MISCELLANEOUS    EXERCISES    FOR    THE    SLATE  381 

162.  "Required,  the  area  of  a  triangle  whose  base  is  46  feet,  and 
perpendicular  heigh'h  32  feet?  Ans.  736  sq.  ft. 

163.  Required — the  area  of  a  triangle  whose  sides  are  19^^,  21, 
and  22i-  ft.  respectively?  Ans.   189  sqf  ft. 

164.  If  the  wheel  of  a  gig  are  4  ft.  7  in.  in  diameter,  how  many 
times  will  they  revolve  in  going  1  mile?  Ans.  366.692+. 

165.  If  the  propelling  wheels  of  a  locomotive  are  3  ft.  4  in.  in  di- 
ameter, and  if  they  make  400  revolutions  per  minute,  how  far  will 
the  engine  move  forward  in  an  hour?        Ans.  47  miles,  192  rods. 
^'166.  A  circular  meadow  I  have  on  my  land, 

-'■  Which  contains  just  two  acres  and  3  tenths  of  ground, 

How  long  must  that  line  be,  which  fixed  to  a  pole, 
Will  just  let  my  horse  gn^ze  nround  on  the  whole? 

Ans."  10.823+ rods=178  ft.,  69.64  in. 
167.  Required — the  solid  contents  of  a  triangular  prism,  the  sides 
of  which  are  4,  5,  and  6  feet,  and  its  heighth  12  feet? 

Ans.   119.0588+  s.  ft. 
%''168.  Required — the  solid  contents  of  a  cylinder  15  ft.  long,  and  3 
ft.  in  diameter?  Ans.   106.029  s.  ft. 

169.  Required — the  area  of  the  surface  and  base  of  a  triangular 
prism,  15  ft.  in  length,  and  measuring  4  ft.  on  each  side? 

Ans.   110.851  sq.  ft. 

170.  Required — ^the  area  of  a  curved  surface  of  a  cylinder  16  ft. 
in  length,  and  3  ft.  in  diameter?  Ans.   150.7*968  sq,  ft. 

171.  What  is  the  area  of  the  surface  of  a  triani^ular  pyramid,  23 
ft,  in  heighth,  and  each  side  of  the  base  being  9  feet? 

Ans,  337  sq.  ft.  72  sq.  in. 

172.  On  the  fourth  of  July  a  pole  was  erected, 
Composed  of  three  pieces  all  nicely  connected  ; 
Five  feet  and  'our  inches  it  measured  arounrl, 

At  the  place  where  it  stood  at  the  top  of  the  ground. 
•  Its  shape  w  s  a  cone,  its  surface  complete. 

And  the  height  of  the  same  was  twice  60  feet. 

How  many  yards  of  blue  ribbon,  procured  at  the  shop, 

Will  wind  round  this  pole  from  bottom  to  top, 
•  Laying  smooth  and  plain  to  be  seen, 

By  leaving  a  space  of  6  inches  between? 

Ans.  214.268. 

173.  A  gentleman  has  a  vessel  in  the  shape  of  a  cone,  and  wishes 
to  know  how  many  gallons  of  wine  it  will  contain,  it  being  25  inches 
deep,  and  5  inches  in  diameter  at  the  largest  end.  Can  you  tell  the 
the  number?  Ans.  708|-=2  qts.  1  pt.  2|  gi, 

174.  How  many  square  inches  of  leather  will  it  require  to  cover 
a  ball  5  inches  in  diameter?  Ans  78.54. 

175.  Required — the  solid  contents  of  a  globe  15  inches  in  diame- 
ter? An«.   1767.16  8.  in. 


382  CX)MMON    ARn'HMKTlC.  ScCt.    XVIII. 

176.  A  stone  was  put  into  a  gallon  measure,  which  was  then 
filled  with  1  qt.  of  water*  Required — ^the  solid  contents  of  the 
stone?  Ans.   173^^  s.  in. 

177.  How  many  bushels  will  a  box  contain  that  is  5  ft.  long,  4  ft. 
wide,  and  4  ft,  deep?  Ans.  64,. 

178.  If  a  cord  that  passes  over  4  movable  pulleys,  be  attached 
to  an  axle  3  inches  in  diameter,  and  the  wheel  is  6  feet  in  "diameter, 
what  weight  can  be  raised  by  the  pulleys,  by  applying  250  lbs.  to 
the  wheel?  Ans.  48000  lbs. 

179.  What  weight  may  be  supported  by  a  screw,  the  threads  of 
which  are  1  inch  apart,  and  the  length  of  the  lever  9  feet,  by  a 
power  of  180  lbs.,  supposing  |-  of  the  power  to  be  lost  in  conse- 
quence of  friction?  Ans.  81430.272  lbs. 

180.  If  a  wheel  containing  84  cogs  runs  into  another  containing 
14  cogs,  and  at  the  other  end  of  the  shaft  of  the  latter,  a  wheel  of 
64  cogs  juts  into  one  of  12  cogs,  which  is  attached  to  a  shaft,  on 
the  other  end  of  which  a  wheel  of  62  cogs  runs  into  one  of  8  cogs, 
that  is  connected  by  a  rod  with  a  band- wheel  32  inches  in  diameter, 
and  the  diameter  of  the  cylinder  where  the  band  runs  is  4  inches. 
How  many  times  does  the  cyhnder  revolve  whilst  the  master-wheel 
revolves  once?  Ans.   1664  times. 

181.  A  and  B  commence  trading  with  $1000  capital,  of  which 
A  furnished  a  part,  and  B.  the  rest.  They  gained  25  per  cent,  and 
shared  it  according  to  the  stock  each  furnished,  when  A  said  to  B — 
"I  have  made  a  handsome  speck."  "Yes,"  said  B,  **but  if  I  only 
had  as  many  such  sums  as  I  now  have  as  you  have  dollars,  I  should 
then  have  ^375. 000.     What  share  of  the  capital  did  each  furnish? 

Ans.  A,  $600,  B,  $400. 

1 82.  A  certain  pole  being  broken  off  by  the  wind,  the  top  fell  27 
feet  from  the  foot  of  the  pole:  the  difference  between  the  length  of 
the  two  pieces  was  9  feet.     Required — the  length  of  the  pole? 

Ans.  81  ft.    ^ 

183.  Two  men,  A  and  B,  are  to  dig  100  rods  of  ditch  for  $100, 
of  which  each  receives  $50.  A  begins  at  one  end,  and  B  bemns  at 
the  other  end.  Upon  settlement,  owing  to  a  difference  in  the  hard- 
ness of  the  ground,  A  receives  $0.25  per  rod  more  than  B.  How 
many  rods  does  each  dig,  and  at  what  price  per  rod? 

.         {  A  digs  43. 845  rods,  at  $1 .  14  per  rod; 
•  I  B  digs  56. 155  rods,  at  $0. 89  per  rod. 

184.  A,  B,  C,  and  D,  agree  to  divide  $120  between  them,  giving 
A  J,  B  J:,  C  i-,  and  D  i-.     Required — the  share  of  each? 

Ans.  $A,  42fV;  B,  $31^;  C,  $25yV;  and  D,  $21yV. 

1 85.  A  gentleman  wishing  to  distribute  some  money  among  some 
children,  ft  und  he  wanted  8  cents  to  give  them  3  cents  apiece;  he 
therefore  gave  them  2  cents  apiece,  and  had  3  cents  left.  How 
many  children  were  there?  Ans    11. 


MISCELLANEOUS    EXERCISES    FOB   TH«    SLATiE.  383 

186,  A  sum  of  money  is  required  to  be  divided  between  A,  B, 
C,  and  D,  in  such  a  manner,  that  as  often  as  A  gets  $2,  B  gets  $3; 
and  B  gets  #4  as  often  as  D  gets  $5;  and  D  gets  $9  as  often  as  C 
gets  ^8;  C.  gets  ^400.  Required — the  share  of  the  otliers,  and 
the  sum  divided? 

D's  share,  8450: 


Sum  divided,  -$1450. 


»  A        S  ^'s  share,  $240; 

'  ^''^'  ^B's  share,  $360; 

187.  A  farmer  observed  that  \  of  his  land  was  sowed  with  wheat, 
j,of  the  remainder  with  oats,  and  jj  of  the  remainder  was  p' anted 
with  corn :  likewise,  he  observed  that  his  wood-land  was  but  j  of 
his  meadow  land,  which  together  amounted  to  70  acres,  and  this 
completed  his  farm.  How  many  acres  had  he  in  all,  and  how  many 
acres  had  he  of  each  crop? 


Of  corn,  50  acres. 

Of  meadow,  50     •' 
Of  timber,      20     " 


Entire  farm,  200  acres. 
Ans.  ^  Of  wheat,       50     " 
(  Of  oats,  30     " 

188.  In  turning  a  gig  within  a  certain  diameter,  it  was  discovered 
that  the  outer  wheel  turned  thrice,  whilst  the  inner  wheel  turned 
but  tw  ce  :  now  supposing  ihe  axletree  to  be  4  ft.  long,  and  the 
wheels  of  an  e  ,ual  size,  the  circumference  described  by  each  wheel 
is  required? 

Ans.  Outer  circumference,  76.3984  ft;  Inner  cir.  50.2656. 

189.  A  father  left  $2000  to  be  divided  between  his  three  so  is, 
aged  14,  16.  and  18  years,  in  such  a  manner  that  the  share  of  each, 
being  plac  d  at  simple  interest,  at  6  per  cent,  until  he  arrived  at  the 
age  of  21  years,  should  amount  to  the  same  sum.  How  much  was 
the  share  of  each? 

i  Share  of  the  youngest,  $606,851+. 
Ans.  <      "     **     "    second,  $662,868+. 
(      "     "     *'    eldest,  $730,279+. 

190.  Four  persons,  A,  B,  C,  and  D,  start  from  the  same  point, 
to  travel  in  the  same  direction  around  an  island  Qb  miles  in  compass: 
A  goes  2  miles,  B  4  miles,  C  6  miles,  and  D  8  miles  an  hour.  How 
long  before  they  next  come  together? 

Ans.   1  day,  8  hours,  30  minutes. 

191.  In  performing  a  journey,  A  has  15  days  the  start  of  B,  and 
travels  7  days  per  week,  whilst  B,  stopping  Sundays,  travels  but  6; 
but  B  travels  as  far  in  4  days  as  A  does  in  5.  .How  many  days  be- 
fore B  overtakes  A?  Ans.  60. 

192.  A,  B,  and  C  can  do  a  piece  of  work  in  6  days  ;  B,  C,  and 
D,  in  7  days;  C,  D,  and  A,  in  1\  days;  and  D,  A,  and  B,  in  9 
days.  In  what  time  can  it  be  done  by  all  of  them  together,  and  by 
each  of  them  singly? 

.        ^  By  all  in  5li4  da.;  Bv  A  in  23^^^  da.;  By  B  in  \%l  da.; 
■^'^^-  I        By  C  in  i3||  da.;' By  D  in  55^^  da. 

193.  A  certain  cellar  was  dug,  the  width  of  which  was  j  of  the 


384  COMMON    ARITHMETIC.  ScCt,    XIX. 

length,  and  the  depth  was  ~  of  the  width  :   160  cubic  yards  of  earth 
was  removed.     Required — the  len-^th  of  the  ditch?     Ans.  30  ft. 

194.  A  in  a  scuffle  seized  on  f-  of  a  parcel  of  sugar  plums  ;  B 
catched  ~  of  them  out  of  his  hands,  and  C  held  on  to  -^  more  ;  D 
ran  off  with  all  that  A  had  left,  except  \,  which  E  afterwards  se- 
cured slyly  for  himself:  next  A  and  C  jointly  set  upon  B,  who  in 
the  conflict  let  fall  \  he  had,  which  was  equally  picked  up"by  D  and 
E:  B  then  knocked  down  C's  hat,  and  to  work  they  went  anew  for 
what  it  contained,  of  which  A  got  \,B,  ~^,  D,  y,  and  C  and  E  equal 
shares  of  what  was  left  of  that  stock :  D  then  struck  J  of  what  A 
and  B  last  acquired  out  of  their  hands,  when  they  with  difficulty 
recovered  -|  of  it  in  equal  shares  again,  but  the  other  three  carried 
off  J  apiece  of  the  same.  Upon  this  they  called  a  truce,  and  agreed 
that  the  -■  of  the  whole  left  by  A  at  first  should  be  equally  divided 
among  them.  What  share  of  the  prize  after  this  distribution  re- 
mained with  each  of  the  competitors?  And  supposing  the  least 
common  multipl  e  of  the  denominators  of  the  fractions  expressing 
their  shares  to  be  the  number  of  sugar  plums  divided,  how  many 
did  each  finally  obtain? 

Ans.totheiast.     ^  A'    2863;    B,    633S;    C,    2438;    D,    10294; 

(  hi,    4951), 

195.  A  merchant  bought  some  cloth  at  the  rate  of  $4  per  yard, 
and  afterwards  bought  ~  as  much  more  at  86  per  yard:  he  sold  it 
all  at  the  rate  of  3  yds.  for  $\o,  and  gained  $140.  How  many 
yards  did  he  buy  in  ail?  Ans.   150. 

196.  A  man  bought  appli^s  at  7  cents  a  dozen,  ~  of  which  he  ex- 
changed for  lemons,  at  the  rate  of  7  apples  for  3  lemons  :  he  then 
sold  aj]  his  apples  and  lemons  at  1^  cents  apiece,  and  gained  36 
cents.     How  many  apples  did  he  buy,  and  what  did  they  cost? 

Ans.  He  bought  7  doz.,  and  they  cost  $0.49. 

197.  The  sum  of  two  numbers  is  812^,  and  4  times  the  less,  is 
equal  to  ~  of  the  greater.     Required — the  numbers? 

Ans.  750  and  62\. 

198.  A  man  bought  6  oranges  and  4  lemons  for  42  cents,  and  at 
the  same  rate,  bought  6  oranges  and  8  lemons  for  49  cents.  Re- 
quired— the  price  of  an  orange,  and  of  a  lemon? 

Ans.  Price  of  an  orange  5  cents,  and  of  a  lemon  3  cents. 

199.  Three  gentlemen  contribute  $675  to  build  a  church,  which 
is  situaed  at  the  distance  of  2  mile-  from  the  first,  2  J  miles  from 
the  second,  and  3^-  miles  from  the  third,  and  agreed  that  their  shares 
should  be  reciprocally  proportional  to  their  distances  from  the 
church.     Required — the  amount  each  contributed? 

.         rrhe  first,  $289.80;  the  second,  $201.60; 
^"^^^  )      the  third,  $165.60. 

200.  A  teacher  agreed  to  teach  a  certain  time  upon  these  condi- 
i  ins  :  Tf  he  had  20  scholars,  ho  was  to  recmve  $26 ;  but  if  he  had 


MISCELLANEOUS    EXERCISES    FOR    THE    SLATE.  385 

30  scholars,  he  was  to  receive  but  -^30.     He  had  29  scholars.     Re- 
quired— his  wages.  Ans.  $29,726.      | 

201.  •*  Between  Sing-Sing  and  Tarry-Town,  I 

I  met  my  worthy  friend  John  Brown  / 

And  seven  daughters,  riding  nags,  ' 

And  every  one  had  seven  bags ; 

In  every  bag  were  thirty  cats, 

And  every  cat  had  forty  rats. 

Besides  a  brood  of  fifty  kittens. 

All  hut  the  nags  were  wearing  mittens  ! 
Mittens,  kittens — cats,  rats — bags,  nags — Browns, 
How  many  were  met  between  the  towns  ? 

"Ans.  2184192." 

Note. — This  question  is  to  be  taken  in  its  plain,  literal,  common-sens© 
meaning,  without  any  quibble,  and  the  niittens  are  to  be  counted  se/jarafcZi/, 
aud  not  in  pairs. 

202.  A  man  spent  $12  for  apples  and  peaches,  giving  $1  for  a 
bushel  of  apples,  and  $1  for  3  bushels  of  peaches  :  he  afterwards 
sold  \  of  his  apples,  and  -J  of  his  peaches  for  $4,  which  wss  $0.50 
more  than  they  cost  him.     How  many  bushels  did  he  buy  of  each? 

Ans.  24  bushels  of  apples;  18  bushels  of  peaches. 


THE   EKD. 


<^ 


?/  ^^^t^/:^^-"^-^/^*^^/ 


r 


ERRATA 


Mr 


6iJ 


'age       5,  Obs.,  4th  line  from  top,  for  '-left/'  read  ''right.** 
46,  Rule  II.,  last  line,  for  "of,"  read  **as." 
57,  and  58,  The  remarks  under  Obs.  20  relate  to  "Practi- 
cal," and  not  to  "Pure  Arithmetic." 
"     122,  Second  line  from  bottom,  for  "IfS-,"  read  'M  or  |J-.'* 
"     132,    "c"  should  read,   "mills  are  reduced  to  cents  by  point- 
ing off  one  figure  to  the  right,  and  to  dollars  by  pointing  off 
three  figures,''  &c. 
"     139,  Third  and  fourth  lines  from  the  bottom,  for  "Art.   1," 

read  "Art.  7." 
"     14",  Fourth  line  from  the  bottom,  for"Art.  1,"  read  "Art.  7." 
«'     155,  Rem.  3,  for  "8  lbs.,"  read  "80  lbs." 
"     156,  Rem.,  for  "Liquor,"  read  "Liquid." 
"     158,  Note  at  the  bottom  of  the  page,  for  "69^,"  read  "69t-V-" 
"     161,  Table,  for  "160  Sq:  Yds.,"-read  '^160  Sq.  rds." 
'•     172,  Ex.  46,  for  "post,"  read  "foot."     " 
«'     173,  Ex.  57,  for  "area,"  read  "arc." 
"     182,  Ex.  14,  for  "Sq.  ft,"  "Sq.  in.,"  read  "s.  ft.  s.  in." 
"     185,  Bottom  of  the*pa^e,  for  "2d,^'  read  "22d." 
"     186,  Obs.  1,  for  for  "Sect.  XIV.,"  read  "Sect.  XIII.;"  and 

Ex.  25,  for  "Area,"  read  "Arc." 
"     223,  Obs.  14,  for  "Ratio  of  Inequality,"  read  "Ratio  of  Les- 
ser equality." 
"  .  226,   Obs.  11,  Illustration,  for  "two  Means,"  read  "two  Ex- 
tremes;" and  for  "two  Extremes,"  read  "two  Means." 
"     227,  Obs."  14,  Rem.,  for   "fir^  terms,"   read  "first  three 

terms." 
"     230,  Fourth  line  above  Obs.  20,  for  "Obs  4,"   read  "Obs. 

14." 
"     237,  Top  line,  for  "17,"  read  "19." 
"     263,  Obs.  8,  Rem.  5,  for  "Obs.  21,"  read  "Obs.  5." 
"     272,  Case  4,  for  "any  number,"  read  "any  sum;"  and  Gen- 
eral Rule,  for  "given  number,"  read  "given  time." 
"     297,  Note,  for  "Obs.  30,"  read  "Obs.  31." 
"     2cJ8,   Obs.  3.3,  for  "Obs.  30,"  read  "Obs.  31;"   and  Ex.   12. 
for  "Obs.  31,"  read  "Obs.  39." 


388  ERRATA. 

"     299,   Obs.  2,  for  "Obs.   15,"  read  ♦'Obs.  17,"  and  for  "Obs. 

18,"  read  "Obs.  20." 
301,  Rem.  1,  for  ''Obs.  15."  read  **Obs.  17." 
303,  Second  line  from   top,  for  "Obs.  18,"  read  "Obs.  20." 
309,  Third  line   from  the  top,  for  "cube  of  8,"  read  "cube 

root  of  8." 
311,  Obs.   15,  for  "their  Factors  are,"  read  "their  Squares 

are." 

321,  Fourth  line  from  bottom,  for  "Obs.  15,"  read  "Obs.  17." 

322,  Operation,  for  "Obs.  15,"  read  "Obs.  17." 
324,  Definition,   for  "Obs.   14,"  read  "Obs.   16;"  and  Cbs. 

2,  for  "Obs.  15,"  read  "Obs.  17." 

326,  Operation,  for  "Obs.  24,"  read  "Obs.  23." 

331,  Second  line  from  top,  for  "Obs.  8,"  read  "Obs.  10." 

338,  Operation,  for  ''J512/'  read  "^512." 

339,  Operation,  for  "Art.  1,  Obs.  2,"  read  "Art.  1,  Obs.  4." 

340,  Obs.  2,  Demonstration,  for  "Obs.   9,"  read  "Obs.  11." 
342,  Obs.   10„   Rem.,  for  "Obs.  17,"  read  "Obs.  19;"  and 

Obs.  12,  f  r  "Obs.  18,"  read  "Obs.  20." 
344,  Obs.  17,  Demonstration,  for  "Obs.  12,"  read  "Obs.  14." 
349,  Bottom  line,  for  "Obs.  11,"  read  "Obs.  14,  a." 
359,  Note  3,  should  be  signed  "Peter  Trueman, 

"Thomas  Jones." 

These  are  the  most  prominent  errors  detected.  Some  few  typo^ 
graphical  errors  have  been  noticed,  which  it  is  not  necessary  to  men- 
tion here.  There  may  perhaps  be  a  few  errors  of  importance  that 
have  escaped  notice.     All  such  will  be  corrected  in  future  editions, 


J  860  17 


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•'*. 


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